6.6 Sketching graphs | Differential calculus

6.5 Second derivative

6.7 Applications of differential calculus

6.6 Sketching graphs (EMCHB)

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Functions of the form \(y = ax^{3} + bx^{2} + cx + d\) (EMCHC)

The effects of \(a\) on a cubic function

Complete the table below and plot the graphs of \(f(x)\) and \(g(x)\) on the same system of axes.

Be careful to choose a suitable scale for the \(y\)-axis.

\[f(x) = 2x^{3} - 5x^{2} - 14x + 8 \qquad \qquad g(x) = -2x^{3} + 5x^{2} + 14x - 8\]

\(x\)	\(-\text{3}\)	\(-\text{2}\)	\(-\text{1}\)	\(\text{0}\)	\(\text{1}\)	\(\text{2}\)	\(\text{3}\)	\(\text{4}\)	\(\text{5}\)
\(f(x)\)							\(-\text{25}\)
\(g(x)\)							\(\text{25}\)

The effects of \(a\) on a cubic graph

Intercepts (EMCHD)

Number of intercepts

Complete the table:

	\(y = x + 1\)	\(y = x^{2} - x - 6\)	\(y = x^{3} + x^{2} - 26x + 24\)
Degree of function
Type of function
Factorised form
No. of \(x\)-intercepts
No. of \(y\)-intercepts

Worked example 17: Determining the intercepts

Given \(f\left(x\right)=-{x}^{3} + 4{x}^{2} + x - 4\), find the \(x\)- and \(y\)-intercepts.

Determine the \(y\)-intercept

The \(y\)-intercept is obtained by letting \(x = 0\):

\begin{align*} y &= -(0)^{3} + 4(0)^{2} + (0) - 4 \\ &= -4 \end{align*}

This gives the point \((0;-4)\).

Use the factor theorem to factorise the expression

We use the factor theorem to find a factor of \(f(x)\) by trial and error:

\begin{align*} f(x) &= -x^{3} + 4x^{2} + x - 4 \\ f(1) &= -(1)^{3} + 4(1)^{2} + (1) - 4 \\ &= 0 \\ \therefore (x - 1) & \text{ is a factor of } f(x) \end{align*}

Factorise further by inspection:

\begin{align*} f(x) &= (x - 1)(-x^{2} + 3x + 4) \\ &= - (x - 1)(x^{2} - 3x - 4) \\ &= - (x - 1)(x + 1)(x - 4) \end{align*}

The \(x\)-intercepts are obtained by letting \(f(x) = 0\):

\begin{align*} 0 &= - (x - 1)(x + 1)(x - 4) \\ \therefore x = -1, x = 1 & \text{ or } x = 4 \end{align*}

This gives the points \((-1;0)\), \((1;0)\) and \((4;0)\).

Intercepts

Textbook Exercise 6.7

Determine the \(x\)- and \(y\)-intercepts of \(f(x)\).

For the \(y\)-intercept, let \(x = 0\):

\begin{align*} f(0) &= (0)^{3} + (0)^{2} - 10(0) + 8 \\ &= 8 \end{align*}

This gives the point \((0;8)\).

We use the factor theorem to find a factor of \(f(x)\) by trial and error:

\begin{align*} f(x) &= x^{3} + x^{2} - 10x + 8 \\ f(1) &= (1)^{3} + (1)^{2} - 10(1) + 8 \\ &= 0 \\ \therefore (x - 1) & \text{ is a factor of } f(x) \end{align*}

Factorise further by inspection:

\begin{align*} f(x) &= (x - 1)(x^{2} + 2x - 8) \\ &= (x - 1)(x - 2)(x + 4) \end{align*}

For the \(x\)-intercept, let \(f(x) = 0\):

\begin{align*} 0 &= (x - 1)(x - 2)(x + 4) \\ \therefore x = 1, x = 2 & \text{ or } x = -4 \end{align*}

This gives the points \((-4;0)\), \((1;0)\) and \((2;0)\).

Draw a rough sketch of the graph.

Is the function increasing or decreasing at \(x = -5\)?

Increasing

\(y = -x^{3} - 5x^{2} + 9x + 45\)

For the \(y\)-intercept, let \(x = 0\):

\begin{align*} y &= -(0)^{3} - 5(0)^{2} + 9(0) + 45 \\ &= 45 \end{align*}

This gives the point \((0;45)\).

We use the factor theorem to find a factor by trial and error:

\begin{align*} \text{Let } f(x) &= -x^{3} - 5x^{2} + 9x + 45 \\ f(3) &= -(3)^{3} - 5(3)^{2} + 9(3) + 45 \\ &= 0 \\ \therefore (x - 3) & \text{ is a factor of } f(x) \end{align*}

Factorise further by inspection:

\begin{align*} f(x) &= (x - 3)(-x^{2} - 8x - 15) \\ &= -(x - 3)(x^{2} + 8x + 15) \\ &= -(x - 3)(x + 3)(x + 5) \end{align*}

For the \(x\)-intercept, let \(y = 0\):

\begin{align*} 0 &= -(x - 3)(x + 3)(x + 5) \\ \therefore x = 3, x = -3 & \text{ or } x = -5 \end{align*}

This gives the points \((-5;0)\), \((-3;0)\) and \((3;0)\) .

\(y = x^{3} - \frac{5}{4}x^{2} - \frac{7}{4}x + \frac{1}{2}\)

For the \(y\)-intercept, let \(x = 0\):

\begin{align*} y &= (0)^{3} - \frac{5}{4}(0)^{2} - \frac{7}{4}(0) + \frac{1}{2} \\ &= \frac{1}{2} \end{align*}

This gives the point \((0;\frac{1}{2})\).

We use the factor theorem to find a factor by trial and error:

\begin{align*} \text{Let } f(x) &= x^{3} - \frac{5}{4}x^{2} - \frac{7}{4}x + \frac{1}{2} \\ f(-1) &= (-1)^{3} - \frac{5}{4}(-1)^{2} - \frac{7}{4}(-1) + \frac{1}{2} \\ &= -1 - \frac{5}{4} + \frac{7}{4} + \frac{1}{2} \\ &= -\frac{4}{4} - \frac{5}{4} + \frac{7}{4} + \frac{2}{4} \\ &= 0 \\ \therefore (x + 1) & \text{ is a factor of } f(x) \end{align*}

Factorise further by inspection:

\begin{align*} f(x) &= (x + 1)(x^{2} - \frac{9}{4}x + \frac{1}{2}) \\ &= (x +1)(x - 2)\left(x - \frac{1}{4}\right) \end{align*}

For the \(x\)-intercept, let \(y = 0\):

\begin{align*} 0 &= (x +1)(x - 2)\left(x - \frac{1}{4}\right) \\ \therefore x = -1, x = 2 & \text{ or } x = \frac{1}{4} \end{align*}

This gives the points \((-1;0)\), \(\left(\frac{1}{4};0\right)\), \((2;0)\) and \(\left(0;\frac{1}{2}\right)\).

\(y = x^{3} - x^{2} - 12x + 12\)

For the \(y\)-intercept, let \(x = 0\):

\begin{align*} y &= (0)^{3} - (0)^{2} - 12(0) + 12\\ &= 12 \end{align*}

This gives the point \((0;12)\).

We use the factor theorem to find a factor by trial and error:

\begin{align*} \text{Let } f(x) &= x^{3} - x^{2} - 12x + 12 \\ f(1) &= (1)^{3} - (1)^{2} - 12(1) + 12 \\ &= 0 \\ \therefore (x - 1) & \text{ is a factor of } f(x) \end{align*}

Factorise further by inspection:

\begin{align*} f(x) &= (x - 1)(x^{2} - 12) \\ &= (x - 1)(x - \sqrt{12})(x + \sqrt{12}) \end{align*}

For the \(x\)-intercept, let \(y = 0\):

\begin{align*} 0 &= (x - 1)(x - \sqrt{12})(x + \sqrt{12}) \\ \therefore x = 1, x = \sqrt{12} & \text{ or } x = -\sqrt{12} \end{align*}

This gives the points \((1;0)\), \((\sqrt{12};0)\) and \((- \sqrt{12};0)\) .

\(y = x^{3} - 16x\)

For the \(y\)-intercept, let \(x = 0\):

\begin{align*} y &= (0)^{3} - 16(0) \\ &= 0 \end{align*}

This gives the point \((0;0)\).

We take out a common factor of \(x\) and then factorise the difference of two squares:

\begin{align*} y &= x(x^{2} - 16) \\ &= x(x - 4)(x + 4) \end{align*}

For the \(x\)-intercept, let \(y = 0\):

\begin{align*} 0 &= x(x - 4)(x + 4) \\ \therefore x = 0, x = 4 & \text{ or } x = -4 \end{align*}

This gives the points \((0;0)\), \((-4;0)\) and \((4;0)\) .

\(y = x^{3} -5x^{2} + 6\)

For the \(y\)-intercept, let \(x = 0\):

\begin{align*} y &= (0)^{3} -5(0)^{2} + 6 \\ &= 6 \end{align*}

This gives the point \((0;6)\).

We use the factor theorem to find a factor by trial and error:

\begin{align*} f(x) &= x^{3} -5x^{2} + 6 \\ f(-1) &= (-1)^{3} -5(-1)^{2} + 6 \\ &= -1 - 5 + 6 \\ &= 0 \\ \therefore (x + 1) & \text{ is a factor of } f(x) \end{align*}

Factorise further by inspection:

\begin{align*} f(x) &= (x + 1)(x^{2} - 6x + 6) \end{align*}

For the \(x\)-intercept, let \(y = 0\):

\begin{align*} 0 &= (x + 1)(x^{2} - 6x + 6) \\ \therefore x = -1, & \text{ or } x = \frac{-(-6) \pm \sqrt{6^{2} - 4(1)(6)}}{2(1)} \\ & = \frac{6 \pm \sqrt{36-24}}{2} \\ & = \frac{6 \pm \sqrt{12}}{2} \\ & = \frac{6 \pm 2\sqrt{3}}{2} \\ & = 3 \pm \sqrt{3} \end{align*}

This gives the points \((-1;0)\), \(( 3 - \sqrt{3};0)\) and \(( 3 + \sqrt{3};0)\) .

Determine all intercepts for \(g(x) = x^{3} + 3x^{2} - 10x\) and draw a rough sketch of the graph.

For the \(y\)-intercept, let \(x = 0\):

\begin{align*} y &= (0)^{3} + 3(0)^{2} - 10(0) \\ &= 0 \end{align*}

This gives the point \((0;0)\).

\begin{align*} y &= x^{3} + 3x^{2} - 10x \\ &= x(x^{2} + 3x - 10) \\ &= x(x + 5)(x - 2) \end{align*}

For the \(x\)-intercept, let \(y = 0\):

\begin{align*} 0 &= x(x + 5)(x - 2) \\ \therefore x = 0, x = -5 & \text{ or } x = 2 \end{align*}

This gives the points \((0;0)\), \((-5;0)\) and \((2;0)\) .

Stationary points (EMCHF)

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Complete the table below for the quadratic function \(f(x)\):

\begin{align*} f(x) &= x^{2} + 2x + 1 \\ f'(x) &= \ldots \ldots \ldots \end{align*}

\(x\)-value	\(-\text{3}\)	\(-\text{1}\)	\(\text{0}\)	\(\text{1}\)	\(\text{3}\)
Gradient of \(f\)
Sign of gradient
\(\begin{array}{c@{\;}c@{\;}l} \text{Increasing function } \left(\nearrow\right) & & \\ \text{Decreasing function } \left(\searrow\right) & & \\ \text{Maximum TP } \left(\cap\right) && \\ \text{Minimum TP } \left(\cup\right) && \end{array}\)

Use the table to draw a rough sketch of the graph of \(f(x)\).
Solve for \(x\) if \(f'(x) = 0\).
Indicate solutions to \(f'(x) = 0\) on the graph.

Complete the table below for the cubic function \(g(x)\):

\begin{align*} g(x) &= 2x^{3} + 3x^{2} -12x \\ g'(x) &= \ldots \ldots \ldots \end{align*}

\(x\)-value	\(-\text{3}\)	\(-\text{2}\)	\(\text{0}\)	\(\text{1}\)	\(\text{3}\)
Gradient of \(g\)
Sign of gradient
\(\begin{array}{c@{\;}c@{\;}l} \text{Increasing function } \left(\nearrow\right) & & \\ \text{Decreasing function } \left(\searrow\right) & & \\ \text{Maximum TP } \left(\cap\right) && \\ \text{Minimum TP } \left(\cup\right) && \end{array}\)

Use the table to draw a rough sketch of the graph of \(g(x)\).
Solve for \(x\) if \(g'(x) = 0\).
Indicate solutions to \(g'(x) = 0\) on the graph.
Complete the following sentence:

The derivative describes the \(\ldots\ldots\) of a tangent to a curve at a given point and we have seen that the \(\ldots\ldots\) of a curve at its stationary point(s) is equal to \(\ldots\ldots\). Therefore, we can use \(\ldots\ldots\) as a tool for finding the stationary points of the graphs of quadratic and cubic functions.

To determine the coordinates of the stationary point(s) of \(f(x)\):

Determine the derivative \(f'(x)\).
Let \(f'(x) = 0\) and solve for the \(x\)-coordinate(s) of the stationary point(s).
Substitute value(s) of \(x\) into \(f(x)\) to calculate the \(y\)-coordinate(s) of the stationary point(s).

Worked example 18: Finding stationary points

Calculate the stationary points of the graph of \(p\left(x\right)= {x}^{3} - 6{x}^{2} + 9x - 4\).

Determine the derivative of \(p\left(x\right)\)

Using the rules of differentiation we get:

\[{p}'\left(x\right)=3{x}^{2} - 12x + 9\]

Let \({p}'\left(x\right)=0\) and solve for \(x\)

\begin{align*} 3{x}^{2} - 12x + 9 & = 0 \\ {x}^{2}-4x+3 & = 0 \\ \left(x-3\right)\left(x-1\right) & = 0 \\ \therefore x = 1 & \text{ or } x = 3 \end{align*}

Therefore, the \(x\)-coordinates of the turning points are \(x=1\) and \(x=3\).

Substitute the \(x\)-values into \(p\left(x\right)\)

We use the \(x\)-coordinates to calculate the corresponding \(y\)-coordinates of the stationary points.

\begin{align*} p\left(1\right) & = {\left(1\right)}^{3}-6{\left(1\right)}^{2} + 9\left(1\right)-4 \\ & = 1 - 6 + 9 - 4\\ & = 0 \end{align*}\begin{align*} p\left(3\right) & = {\left(3\right)}^{3}- 6{\left(3\right)}^{2} + 9\left(3\right)-4 \\ & = 27 - 54 + 27 - 4 \\ & = -4 \end{align*}

Write final answer

The turning points of the graph of \(p\left(x\right)= {x}^{3} - 6{x}^{2} + 9x - 4\) are \(\left(1;0\right)\) and \(\left(3;-4\right)\).

Local maximum and local minimum

We have seen that the graph of a quadratic function can have either a minimum turning point (“smile”) or a maximum turning point (“frown”).

For cubic functions, we refer to the turning (or stationary) points of the graph as local minimum or local maximum turning points. The diagram below shows local minimum turning point \(A(1;0)\) and local maximum turning point \(B(3;4)\). These points are described as a local (or relative) minimum and a local maximum because there are other points on the graph with lower and higher function values.

Stationary points

Textbook Exercise 6.8

Use differentiation to determine the stationary point(s) for \(g(x) = -x^{2}+5x-6\).

\begin{align*} g(x) &= -x^{2}+5x-6 \\ g'(x) &= -2x+5 \\ \text{At stationary point: } y'&=0 \\ -2x+5&=0 \\ -2x&=-5\\ x&=\text{2,5} \\ \text{Substitute } x &=\text{2,5} \\ \therefore y&=-(\text{2,5})^{2} + 5(\text{2,5}) - 6 \\ &=-\text{6,25} + \text{12,5} - 6 \\ &=\text{0,25} \\ \therefore \text{ Stationary point } & \left(\frac{5}{2};\frac{1}{4}\right) \end{align*}

Determine the \(x\)-values of the stationary points for \(f(x) = -\frac{1}{3}x^{3} + \frac{1}{2}x^2 + 6x + 5\).

\begin{align*} f(x) &= -\frac{1}{3}x^{3} + \frac{1}{2}x^2 + 6x + 5 \\ f'(x) &=-x^{2}+x+6 \\ \text{At stationary point: } y'&=0 \\ -x^{2}+x+6 &=0 \\ x^{2}-x-6&=0 \\ (x + 2)(x - 3)&=0 \\ \therefore x = -2 &\text{ or } x = 3 \end{align*}

\(y = (x-1)^{3}\)

\begin{align*} y &= (x-1)^{3} \\ y&=x^{3}-3x^{2}+3x-1 \\ y'&=3x^{2}-6x+3\\ \text{At stationary point: } y'&=0 \\ 3(x^{2}-2x+1)&=0 \\ (x-1)^{2}&=0 \\ \therefore x&=1 \\ \text{Substitute } x&=1 \\ y &= (1-1)^{3} = 0 \\ \therefore \text{ Stationary point }& (1;0) \end{align*}

\(y = x^{3}-5x^{2} + 1\)

\begin{align*} y &= x^{3}-5x^{2} + 1 \\ y'&=3x^{2}-10x\\ \text{At stationary point: } y'&=0 \\ 3x^{2}-10x&=0 \\ x(3x-10)&=0 \\ \therefore x&=0, x=\frac{10}{3} \end{align*} \begin{align*} \text{Substitute } x&=0 \\ y &= (0)^{3}-5(0)^{2}+ 1 = 1 \\ \therefore \text{ Stationary point }&(0;1) \\ \text{Substitute } x&=\frac{10}{3} \\ y &= \left(\frac{10}{3}\right)^{3}-5\left(\frac{10}{3}\right)^{2}+ 1\\ &= \frac{1000}{27} - \frac{500}{9} + 1 \\ &= -\frac{473}{27} \\ \therefore \text{ Stationary point }&\left(\frac{10}{3};-\frac{473}{27}\right) \end{align*}

\(y+7x = 1\)

\begin{align*} y &= - 7 x + 1 \\ y'&=-7 \end{align*}

This is a straight line with a constant gradient, therefore there is no stationary point.

Sketching cubic graphs (EMCHG)

General method for sketching cubic graphs:

Consider the sign of \(a\) and determine the general shape of the graph.
Determine the \(y\)-intercept by letting \(x=0\).
Determine the \(x\)-intercepts by factorising \(a{x}^{3}+b{x}^{2}+cx+d=0\) and solving for \(x\).
Find the \(x\)-coordinates of the turning points of the function by letting \(f'(x) = 0\) and solving for \(x\).
Determine the \(y\)-coordinates of the turning points by substituting the \(x\)-values into \(f\left(x\right)\).
Plot the points and draw a smooth curve.

Worked example 19: Sketching cubic graphs

Sketch the graph of \(g\left(x\right)={x}^{3}-3{x}^{2}-4x\).

Determine the shape of the graph

The coefficient of the \(x^{3}\) term is greater than zero, therefore the graph will have the following shape:

Determine the intercepts

The \(y\)-intercept is obtained by letting \(x = 0\):

\begin{align*} g\left(0\right)&=\left(0\right)^{3}-3\left(0\right)^{2}-4\left(0\right)\\ & = 0 \end{align*}

This gives the point \((0;0)\).

The \(x\)-intercept is obtained by letting \(g(x) = 0\) and solving for \(x\):

\begin{align*} 0 & = {x}^{3}-3{x}^{2}-4x \\ & = x(x^{2} - 3x - 4) \\ & = x(x - 4)(x+1)\\ \therefore x=-1, \enspace & x = 0 \text{ or } x = 4 \end{align*}

This gives the points \((-1;0)\), \((0;0)\) and \((4;0)\).

Calculate the stationary points

Find the \(x\)-coordinates of the stationary points by setting \({g}'\left(x\right)=0\):

\begin{align*} {g}'\left(x\right) & = 3x^{2} - 6x - 4 \\ 0 & = 3x^{2} - 6x - 4 \\ \text{Using the quadratic formula} \enspace x & = \frac{-(-6) \pm \sqrt{(-6)^{2} - 4(3)(-4)}}{2(3)} \\ & = \frac{6 \pm \sqrt{36 + 48}}{6} \\ \therefore x = \text{2,53} & \text{ or } x = -\text{0,53} \end{align*}

Substitute these \(x\)-coordinates into \(g(x)\) to determine the corresponding \(y\)-coordinates:

\begin{align*} g\left(x\right)&=\left(\text{2,53}\right)^{3}-3\left(\text{2,53}\right)^{2}-4\left(\text{2,53}\right)\\ & = -\text{13,13} \end{align*}\begin{align*} g\left(x\right)&=\left(-\text{0,53}\right)^{3}-3\left(-\text{0,53}\right)^{2}-4\left(-\text{0,53}\right) \\ & = \text{1,13} \end{align*}

Therefore, the stationary points are \(\left(\text{2,53}; -\text{13,13}\right)\) and \(\left(-\text{0,53}; \text{1,13}\right)\).

Draw a neat sketch

Concavity

Concavity indicates whether the gradient of a curve is increasing, decreasing or stationary.

Concave up: the gradient of the curve increases as \(x\) increases.
Concave down: the gradient of the curve decreases as \(x\) increases.
Zero concavity: the gradient of the curve is constant.

The diagram below shows the graph of the cubic function \(k(x) = x^{3}\). The first derivative of \(k(x)\) is a quadratic function, \(k'(x) = 3x^{2}\) and the second derivative is a linear function, \(k''(x) = 6x\).

Notice the following:

\(k''(x) > 0\), the graph is concave up.
\(k''(x) < 0\), the graph is concave down.
\(k''(x) = 0\), change in concavity (point of inflection).

Points of inflection

This is the point where the concavity of a curve changes, as shown in the diagram below. If \(a < 0\), then the concavity changes from concave up (purple) to concave down (grey) and if \(a > 0\), concavity changes from concave down (blue) to concave up (green). Unlike a turning point, the gradient of the curve on the left-hand side of an inflection point (\(P\) and \(Q\)) has the same sign as the gradient of the curve on the right-hand side.

A graph has a horizontal point of inflection where the derivative is zero but the sign of the gradient of the curve does not change. That means the graph (shown below) will continue to increase or decrease after the stationary point.

In the example above, the equation \(k'(x) = 3x^{2}\) indicates that the gradient of this curve will always be positive (except where \(x = 0\)). Therefore, the stationary point is a point of inflection.

\(f\): cubic function	\(f'\): quadratic function	\(f''\): linear function
(blue graph)	(green graph)	(red graph)
turning points \(\rightarrow\)	\(x\)-intercepts
point of inflection	\(\leftarrow\) turning point \(\rightarrow\)	\(x\)-intercept

Concavity and points of inflection

Textbook Exercise 6.9

Complete the following for each function:

Determine and discuss the change in gradient of the function.
Determine the concavity of the graph.
Find the inflection point.
Draw a sketch of the graph.

\(f: y=-2x^{3}\)

Gradient: \(f'(x) = -6x^{2}\)

\(f'(x) < 0\) for \(x < 0\): function decreasing

\(f'(x) = 0\) for \(x = 0\): function stationary

\(f'(x) > 0\) for \(x > 0\): function increasing
Concavity: \(f''(x)=-12x\)

\(f''(x) > 0\) for \(x < 0\): concave up

\(f''(x) = 0\) for \(x = 0\): point of inflection

\(f''(x) < 0\) for \(x > 0\): concave down
Point of inflection: \((0;0)\)

\(g(x)=\frac{1}{8}x^{3}+1\)

\(g(x)=y=\frac{1}{8}x^{3}+1\)

Gradient: \(g'(x)=\frac{3}{8}x^{2}\)

Gradient always \(>0\) (since \(\frac{3}{8}x^{2}\geq0\)) for all \(x\) except \(x=0\) where the gradient will be \(\text{0}\).
Concavity: \(g''(x) = \frac{3}{4}x\)

\(g''(x) < 0\) for \(x < 0\): concave down

\(g''(x) = 0\) for \(x = 0\): point of inflection

\(g''(x) > 0\) for \(x > 0\): concave up
Point of inflection: \((0;1)\)

Intercepts:

\begin{align*} y_{\text{int}}: \text{ let } x &= 0\\ \therefore y & = 1 \\ \therefore & (0;1) \end{align*} \begin{align*} y_{\text{int}}: \text{ let } y&=0\\ \frac{1}{8}x^{3}+1&=0\\ x^{3}&=-8 \\ x&=-2 \\ \therefore x_{\text{int}} & (-2;0) \end{align*}

\(h: x \rightarrow (x-2)^{3}\)

\(h(x)=(x-2)^{3}=x^{3}-6x^{2}+12x-8\)

Gradient: \begin{align*} h'(x)&=3x^{2}-12x+12 \\ &=3(x^{2}-4x+4) \\ &= 3(x-2)^{2} \end{align*}

Gradient always \(>0\) for all \(x\) except \(x=2\) where the gradient will be \(\text{0}\).

Stationary point: \((2;0)\)
Concavity: \(h''(x) = 6x-12\)

\(h''(x) < 0\) for \(x < 2\): concave down

\(h''(x) = 0\) for \(x = 2\): point of inflection

\(h''(x) > 0\) for \(x > 2\): concave up
Point of inflection: \((2;0)\)

Intercepts:

\(y_{\text{int}}: (0;-8)\)

\(x_{\text{int}}: (2;0)\)

Interpreting graphs

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Worked example 20: Interpreting graphs

Consider the graph of the derivative of \(g(x)\).

For which values of \(x\) is \(g(x)\) decreasing?
Determine the \(x\)-coordinate(s) of the turning point(s) of \(g(x)\).
Given that \(g(x) = ax^{3} + bx^{2} + cx\), calculate \(a\), \(b\) and \(c\).

Examine the parabolic graph and interpret the given information

We know that \(g'(x)\) describes the gradient of \(g(x)\). To determine where the cubic function is decreasing, we must find the values of \(x\) for which \(g'(x) < 0\):

\[\{x: - 2 < x < 1; x \in \mathbb{R} \} \text{ or we can write } x \in (-2;1)\]

Determine the \(x\)-coordinate(s) of the turning point(s)

To determine the turning points of a cubic function, we let \(g'(x) = 0\) and solve for the \(x\)-values. These \(x\)-values are the \(x\)-intercepts of the parabola and are indicated on the given graph:

\[x = -2 \text{ or } x = 1\]

Determine the equation of \(g(x)\)

\begin{align*} g(x) &= ax^{3} + bx^{2} + cx \\ g'(x) &= 3ax^{2} + 2bx + c \end{align*}

From the graph, we see that the \(y\)-intercept of \(g'(x)\) is \(-\text{6}\).

\begin{align*} \therefore c &= - 6 \\ g'(x) &= 3ax^{2} + 2bx - 6 \\ \text{Substitute } x = -2: \enspace g'(-2) &= 3a(-2)^{2} + 2b(-2) - 6 \\ 0 &= 12a - 4b - 6 \ldots \ldots (1) \\ \text{Substitute } x = 1: \enspace g'(1) &= 3a(1)^{2} + 2b(1) - 6 \\ 0 &= 3a + 2b - 6 \ldots \ldots (2) \end{align*}\begin{align*} \text{Eqn. }(1) - 4 \text{ Eqn. }(2): \quad 0 &= 0 -12b + 18 \\ \therefore b &= \frac{3}{2} \\ \text{And } 0 &= 3a + 2 \left( \frac{3}{2} \right) - 6 \\ 0 &= 3a - 3 \\ \therefore a &= 1 \\ & \\ g(x) &= x^{3} + \frac{3}{2}x^{2} - 6x \end{align*}

Video: 28VW

Mixed exercises on cubic graphs

Textbook Exercise 6.10

Show that \(\left(x-1\right)\) is a factor of \(f\left(x\right)\) and hence factorise \(f\left(x\right)\).

First substitute in \(x=1\) to check if \((x-1)\) is a factor:

\begin{align*} f(1) & = (1)^{3} + (1)^{2} - 5(1) + 3 \\ & = 1 + 1 - 5 + 3 \\ & = 0 \end{align*}

Therefore \((x-1)\) is a factor of \(f(x)\).

\begin{align*} f(x) & = (x-1)(x^{2}+2x-3)\\ & = (x-1)(x+3)(x-1) \end{align*}

Determine the coordinates of the intercepts and the turning points.

For the \(y\)-intercept, let \(x = 0\): \(f(0) = 0^{3} + 0^{2} - 5(0) + 3 = 3\), which gives the point \((0;3)\).

For the \(x\)-intercepts, let \(f(x) = 0\):

\begin{align*} 0 & = (x-1)(x+3)(x-1) \end{align*}

Therefore, the \(x\)-intercepts are: \((1;0)\) and \((-3;0)\).

Find the turning points:

\begin{align*} f'(x) & = 3x^{2} + 2x - 5 \\ 0 & = 3x^{2} + 2x - 5 \\ 0 & = (3x+5)(x-1) \\ \therefore x = -\frac{5}{3} &\text{ or } x = 1 \end{align*}

To find the \(y\)-values, we substitute these two values for \(x\) into the original equation. We already know that when \(x=1\), \(y=0\). Substituting in the other value gives:

\begin{align*} f\left(-\frac{5}{3}\right) & = \left(-\frac{5}{3} \right)^{3} + \left(-\frac{5}{3} \right)^{2} - 5\left(-\frac{5}{3} \right) + 3 \\ & = \frac{256}{27} \end{align*}

Therefore the turning points are: \((1;0)\) and \(\left(-\frac{5}{3};\frac{256}{27}\right)\)

Sketch the graph.

Sketch the graph of \(f\left(x\right)=-{x}^{3}+4{x}^{2}+11x-30\). Show all the turning points and intercepts with the axes.

We find the \(y\)-intercept by letting \(x = 0\):

\begin{align*} f(x) & = -x^{3} +4x^{2} +11x - 30 \\ f(0) & = -(0)^{3} + 4(0)^{2} +11(0) - 30\\ & = -30 \end{align*}

The \(y\)-intercept is: \((0;-30)\)

We find the \(x\)-intercepts by letting \(f(x) = 0\).

We use the factor theorem to check if \((x-1)\) is a factor.

\begin{align*} f(x) & = -x^{3} + 4x^{2} + 11x - 30 \\ f(1) & = -(1)^{3} + 4(1)^{2} + 11(1) - 30 \\ & = -16 \end{align*}

Therefore, \((x-1)\) is not a factor.

We now use the factor theorem to check if \((x+1)\) is a factor.

\begin{align*} f(x) & = -x^{3} + 4x^{2} + 11x - 30 \\ f(-1) & = -(-1)^{3} + 4(-1)^{2} + 11(-1) - 30 \\ & = -36 \end{align*}

Therefore, \((x+1)\) is not a factor.

We now try \((x-2)\):

\begin{align*} f(x) & = -x^{3} + 4x^{2} + 11x - 30 \\ f(2) & = -(2)^{3} + 4(2)^{2} + 11(2) - 30 \\ & = 0 \end{align*}

Therefore, \((x-2)\) is a factor.

\begin{align*} f(x) & = (x-2)(-x^{2} + 2x + 15) \\ & = -(x-2)(x^{2} - 2x - 15) \\ & = - (x - 2)(x + 3)(x - 5) \end{align*}

The \(x\)-intercepts are: \((2;0), (-3;0), (5;0)\).

For the turning points, let \(f'(x)=0\).

\begin{align*} f'(x) & = -3x^{2} + 8x + 11 \\ &= -(3x^{2}-8x-11) \\ \therefore 0 & = (3x-11)(x+1) \end{align*}

The \(x\)-coordinates of the turning points are: \(x=-1\) and \(x=\frac{11}{3}\).

The \(y\)-coordinates of the turning points are calculated as:

\begin{align*} f(-1) & = -(-1)^{3} + 4(-1)^{2} +11(-1) - 30 \\ & = -36 \end{align*}

and

\begin{align*} f(\frac{11}{3}) & = -\left(\frac{11}{3}\right)^{3} + 4\left(\frac{11}{3}\right)^{2} + 11\left(\frac{11}{3}\right) - 30 \\ & = \frac{400}{3} \end{align*}

Therefore, the turning points are: \((-1;-36)\) and \((\frac{11}{3};\frac{400}{27})\).

Given \(g(x)=x^{3}-4x^{2}-11x+30\), sketch the graph of \(g\) without any further calculations. Describe the method for drawing the graph.

It is the mirror image of \(f\) in the \(x\)-axis. In other words, we reflect the graph of \(f\) about the \(x\)-axis.

Find the equation of \(f\).

\(x\)-intercepts are \((2;0)\), \((2;0)\) and \((5;0)\).

\begin{align*} \therefore \text{ Equation of } f: \enspace y&=a(x-2)^{2}(x-5) \\ &=a(x^{2}-4x+4)(x-5) \\ &=a(x^{3} - 9x^{2} + 24x -20) \\ &=ax^{3}-9ax^{2}+24ax-20a \end{align*}

From the graph, \(y_{\text{int}}\) is \((0;20)\).

\begin{align*} \therefore -20a&=-20 \\ a&=1 \\ \therefore f: \enspace y &= x^{3}-9x^{2}+24x-20 \end{align*}

Find the coordinates of turning point \(A\).

The turning points are where \(f'(x)=0\):

\begin{align*} f'(x)&=3x^{2}-18x+24 \\ 3x^{2}-18x+24&=0 \\ (3x-12)(x-2)&=0\\ \therefore 3x-12=0 &\text{ or } x-2 =0 \\ \therefore x = 4 &\text{ or } x=2 \\ \therefore \text{ Point } A \text{ is } (4;y) \end{align*}

Substitute \(x=4\) into the equation of \(f\) to calculate corresponding \(y\)-value:

\begin{align*} y&=(4)^{3}-(9)(4)^{2}+24(4)-20 \\ &=64-144+96-20\\ &=-4 \\ \therefore \text{ Point } A \text{ is } (4;-4) \end{align*}

Find the intercepts and stationary point(s) of \(f(x)=-\frac{1}{3}x^{3}+2\) and draw a sketch of the graph.

\begin{align*} y_{\text{int}}: \text{ Let }x&=0 \\ \therefore y&=-\frac{1}{3}(0)^{3}+2\\ y&=2 \end{align*}

This gives the point \((0;2)\).

\begin{align*} x_{\text{int}}: \text{ Let }y&=0 \\ -\frac{1}{3}x^{3}+2&=0\\ -\frac{1}{3}x^{3} &= - 2\\ x^{3}&=6\\ x &= \sqrt[3]{6} \end{align*}

This gives the point \(\left(\sqrt[3]{6};0\right)\).

\begin{align*} \text{Stationary points: where } f'(x)&=0 \\ f'(x)&=-x^{2} \\ -x^{2}&=0 \\ \therefore x &=0 \\ \text{Substitute into } f \therefore y &=-\frac{1}{3}(0)^{2}+2\\ y&=2 \end{align*}

There is only one stationary point at \((0;2)\). From \(f'(x) = -x^{2}\), we know that the gradient of the function is always negative, therefore this is a point of inflection.

For which values of \(x\) will:

\(f(x)<0\)
\(f'(x)<0\)
\(f''(x)<0\)

Motivate each answer.

For \(f(x)<0\), the function values are negative and this is true for \(x>\sqrt[3]{6}\).
For \(f'(x)<0\), the gradient of \(f(x)\) is negative, so where \(f'(x)<0\), and this is true for \(x \in R\), \(x \ne 0\).
\(f''(x)<0\) where \(f(x)\) is concave down and this is true where \(x > 0\).

\begin{align*} g(-6)&=g(-\text{1,5})=g(2)=0 \\ g'(-4)&=g'(1)=0\\ g'(x)&>0 \text{ for } x <-4 \text{ or } x > 1\\ g'(x)& <0 \text{ for } -4<x<1 \end{align*}

\begin{align*} h(-3)&=0 \\ h(0)&=4\\ h(-1)&=3 \\ h'(-1)&=0 \\ h''(-1)&=0 \\ h'(x)& >0 \text{ for all } x \text{ values except } x=-1 \end{align*}

length \(OB\)

\begin{align*} \text{Find } x_{\text{int}} \text{ of } f \text{: let } y&=0\\ \therefore -(x+2)(x-1)(x-6)&=0\\ x=-2 \enspace (B), \enspace x=1 \enspace (E), \enspace x&=6 \enspace (G) \\ \therefore OB&=\text{2}\text{ units} \quad (\text{Note length is always positive}) \end{align*}

length \(OE\)

\(OE = \text{1}\text{ unit}\)

length \(EG\)

\(EG = 6 - 1 = \text{5}\text{ units}\)

length \(OD\)

Find \(y_{\text{int}}\) by writing \(f\) in expanded form:

\begin{align*} y&=-(x+2)(x^{2}-7x+6)\\ &= -(x^{3}-5x^{2}-8x+12) \\ &= -x^{3} + 5x^{2} + 8x - 12 \\ \therefore OD&= \text{12}\text{ units} \end{align*}

coordinates of \(C\) and \(F\)

\(C\) and \(F\) are the turning points.

To find the \(x\)-coordinates of \(C\) and \(F\), find \(f'(x)\):

\begin{align*} f'(x)&=-3x^{2}+10x+8\\ \text{At turning points, } f'(x)&=0\\ \therefore -3x^{2} +10x +8&=0\\ 3x^{2}-10x-8&=0\\ (3x+2)(x-4)& = 0\\ x=-\frac{2}{3} \text{ or } x&=4 \end{align*}

Substitute to find \(y\)-coordinates of the turning points:

\begin{align*} x = -\frac{2}{3}: \enspace y&=-\left(-\frac{2}{3}\right)^{3} + 5\left(-\frac{2}{3}\right)^{2} + 8\left(-\frac{2}{3}\right)-12\\ &= \frac{8}{27} + \frac{20}{9} - \frac{16}{3} -12 \\ &= \frac{8+60-144-324}{27} \\ &= -\frac{400}{27} \\ &\approx -\text{14,8} \\ \therefore C&= \left(-\frac{2}{3};-\frac{400}{27}\right) \\ x = 4: \enspace y&=-(4)^{3} + 5(4)^{2} + 8(4)-12\\ &= -64 + 80 + 32 -12 \\ &= 36 \\ \therefore F&= \left(4;36\right) \end{align*}

length \(AF\)

\(A\) has same \(y\)-coordinate as \(F\), \(y=36\). Therefore \(-x^{3} + 5x^{2} +8x -12 = 36\) at \(F\).

Solve for \(x\) to find the \(x\)-coordinate of \(A\): \(x^{3}-5x^{2}-8x+48=0\).

We know that \(x=4\) is a solution of this equation, therefore \((x-4)\) is a factor.

\begin{align*} \therefore x^{3}-5x^{2} -8x + 48 &= (x-4)(x^{2}-x-12)=0 \\ \text{Solve } x^{2}-x-12&=0 \\ (x-4)(x+3)&=0 \\ \therefore x=4, \enspace x&=-3 \\ \therefore x\text{-coordinate of } A \text{ is } -3 \\ x\text{-coordinate of } F \text{ is } 4 \\ \therefore AF = \text{7}\text{ units} \end{align*}

average gradient between \(E\) and \(F\)

\begin{align*} \text{Average gradient} &= \frac{f(4)-f(1)}{4-1} \\ &=\frac{36-0}{3} \\ &=12 \end{align*}

the equation of the tangent to the graph at \(E\)

\begin{align*} \text{Gradient at } E&= f'(1) \\ f'(x)&=-3x^{2}+10x+8 \\ f'(1)&=-3(1)^{2} + 10(1)+8 \\ &=15 \\ \therefore y&=15x+c\\ \text{Tangent goes through } (1;0) \\ \therefore 0 &= 15(1) +c\\ \therefore c &= -15 \end{align*}

Therefore the equation of the tangent is \(y=15x-15\).

Given the graph of a cubic function with the stationary point \((3;2)\), sketch the graph of the derivative function if it is also given that the gradient of the graph is \(-5\) at \(x=0\).

Derivative will be of the second degree: parabola. Gradient of function is negative throughout, except at \((3;2)\), where gradient \(=0, \quad \therefore g'(3) = 0\). \(g'\) has a maximum value of \(\text{0}\) where \(x = 3\). Therefore \(g'(0)=-5\).

The sketch below shows the graph of \(h'(x)\) with \(x\)-intercepts at \(-\text{5}\) and \(\text{1}\).

Draw a sketch graph of \(h(x)\) if \(h(-\text{5})=2\) and \(h(1)=6\).

\(h(x)\) is a cubic function because \(h'(x)\) is a parabola. \(h(x)\) has two turning points because \(h'(x)\) has two \(x\)-intercepts.

The \(x\)-values of the turning points of \(h(x)\) are the \(x\)-intercepts of \(h'(x)\), where \(h'(x)=0\).

So \(h(x)\) will have turning points at \((-5;2)\): minimum turning point (where gradient changes from negative to positive) and at \((1;6)\): maximum turning point (where gradient changes from positive to negative).

6.5 Second derivative

Table of Contents

6.7 Applications of differential calculus