5.1 Differentiation (first principles, rules) and sketching graphs (Page 3/3)

Siyavula textbooks: grade 12 Page 3 / 3

\begin{matrix} 2 x & = & 0 \\ x & = & 0 \end{matrix}

We then substitute this into the equation of the graph (i.e. $y = x^{2}$ ) to determine the $y$ -coordinate of the turning point: $f (0) = {(0)}^{2} = 0$ This corresponds to the point that we have previously calculated.

Calculate the turning points of the graph of the function $f (x) = 2 x^{3} - 9 x^{2} + 12 x - 15$ .

Determine the derivative of $f (x)$
Using the rules of differentiation we get: $f^{'} (x) = 6 x^{2} - 18 x + 12$
Set $f^{'} (x) = 0$ and calculate $x$ -coordinate of turning point
$\begin{matrix} 6 x^{2} - 18 x + 12 & = & 0 \\ x^{2} - 3 x + 2 & = & 0 \\ (x - 2) (x - 1) & = & 0 \end{matrix}$

Therefore, the turning points are at $x = 2$ and $x = 1$ .
Substitute $x$ -coordinate of turning point into $f (x)$ to determine $y$ -coordinates
$\begin{matrix} f (2) & = & 2 {(2)}^{3} - 9 {(2)}^{2} + 12 (2) - 15 \\ = & 16 - 36 + 24 - 15 \\ = & - 11 \end{matrix}$

$\begin{matrix} f (1) & = & 2 {(1)}^{3} - 9 {(1)}^{2} + 12 (1) - 15 \\ = & 2 - 9 + 12 - 15 \\ = & - 10 \end{matrix}$
Write final answer
The turning points of the graph of $f (x) = 2 x^{3} - 9 x^{2} + 12 x - 15$ are (2,-11) and (1,-10).

We are now ready to sketch graphs of functions.

Method:

Sketching Graphs: Suppose we are given that $f (x) = a x^{3} + b x^{2} + c x + d$ , then there are five steps to be followed to sketch the graph of the function:

If $a > 0$ , then the graph is increasing from left to right, and may have a maximum and a minimum. As $x$ increases, so does $f (x)$ . If $a < 0$ , then the graph decreasing is from left to right, and has first a minimum and then a maximum. $f (x)$ decreases as $x$ increases.
Determine the value of the $y$ -intercept by substituting $x = 0$ into $f (x)$
Determine the $x$ -intercepts by factorising $a x^{3} + b x^{2} + c x + d = 0$ and solving for $x$ . First try to eliminate constant common factors, and to group like terms together so that the expression is expressed as economically as possible. Use the factor theorem if necessary.
Find the turning points of the function by working out the derivative $\frac{d f}{d x}$ and setting it to zero, and solving for $x$ .
Determine the $y$ -coordinates of the turning points by substituting the $x$ values obtained in the previous step, into the expression for $f (x)$ .
Use the information you're given to plot the points and get a rough idea of the gradients between points. Then fill in the missing parts of the function in a smooth, continuous curve.

Draw the graph of $g (x) = x^{2} - x + 2$

Determine the $y$ -intercept
The $y$ -intercept is obtained by setting $x = 0$ . $g (0) = {(0)}^{2} - 0 + 2 = 2$ The turning point is at (0,2).
Determine the $x$ -intercepts
The $x$ -intercepts are found by setting $g (x) = 0$ .

$\begin{matrix} g (x) & = & x^{2} - x + 2 \\ 0 & = & x^{2} - x + 2 \end{matrix}$

Using the quadratic formula and looking at $b^{2} - 4 a c$ we can see that this would be negative and so this function does not have real roots. Therefore, the graph of $g (x)$ does not have any $x$ -intercepts.
Find the turning points of the function
Work out the derivative $\frac{d g}{d x}$ and set it to zero to for the $x$ coordinate of the turning point. $\frac{d g}{d x} = 2 x - 1$

$\begin{matrix} \frac{d g}{d x} & = & 0 \\ 2 x - 1 & = & 0 \\ 2 x & = & 1 \\ x & = & \frac{1}{2} \end{matrix}$
Determine the $y$ -coordinates of the turning points by substituting the $x$ values obtained in the previous step, into the expression for $f (x)$ .
$y$ coordinate of turning point is given by calculating $g (\frac{1}{2})$ .

$\begin{matrix} g (\frac{1}{2}) & = & {(\frac{1}{2})}^{2} - (\frac{1}{2}) + 2 \\ = & \frac{1}{4} - \frac{1}{2} + 2 \\ = & \frac{7}{4} \end{matrix}$

The turning point is at $(\frac{1}{2}, \frac{7}{4})$
Draw a neat sketch

Sketch the graph of $g (x) = - x^{3} + 6 x^{2} - 9 x + 4$ .

Calculate the turning points
Find the turning points by setting $g^{'} (x) = 0$ .

If we use the rules of differentiation we get $g^{'} (x) = - 3 x^{2} + 12 x - 9$

$\begin{matrix} g^{'} (x) & = & 0 \\ - 3 x^{2} + 12 x - 9 & = & 0 \\ x^{2} - 4 x + 3 & = & 0 \\ (x - 3) (x - 1) & = & 0 \end{matrix}$

The $x$ -coordinates of the turning points are: $x = 1$ and $x = 3$ .

The $y$ -coordinates of the turning points are calculated as:

$\begin{matrix} g (x) & = & - x^{3} + 6 x^{2} - 9 x + 4 \\ g (1) & = & - {(1)}^{3} + 6 {(1)}^{2} - 9 (1) + 4 \\ = & - 1 + 6 - 9 + 4 \\ = & 0 \end{matrix}$

$\begin{matrix} g (x) & = & - x^{3} + 6 x^{2} - 9 x + 4 \\ g (3) & = & - {(3)}^{3} + 6 {(3)}^{2} - 9 (3) + 4 \\ = & - 27 + 54 - 27 + 4 \\ = & 4 \end{matrix}$

Therefore the turning points are: $(1, 0)$ and $(3, 4)$ .
Determine the $y$ -intercepts
We find the $y$ -intercepts by finding the value for $g (0)$ .

$\begin{matrix} g (x) & = & - x^{3} + 6 x^{2} - 9 x + 4 \\ y_{i n t} = g (0) & = & - {(0)}^{3} + 6 {(0)}^{2} - 9 (0) + 4 \\ = & 4 \end{matrix}$
Determine the $x$ -intercepts
We find the $x$ -intercepts by finding the points for which the function $g (x) = 0$ .

$g (x) = - x^{3} + 6 x^{2} - 9 x + 4$

Use the factor theorem to confirm that $(x - 1)$ is a factor. If $g (1) = 0$ , then $(x - 1)$ is a factor.

$\begin{matrix} g (x) & = & - x^{3} + 6 x^{2} - 9 x + 4 \\ g (1) & = & - {(1)}^{3} + 6 {(1)}^{2} - 9 (1) + 4 \\ = & - 1 + 6 - 9 + 4 \\ = & 0 \end{matrix}$

Therefore, $(x - 1)$ is a factor.

If we divide $g (x)$ by $(x - 1)$ we are left with: $- x^{2} + 5 x - 4$ This has factors $- (x - 4) (x - 1)$

Therefore: $g (x) = - (x - 1) (x - 1) (x - 4)$

The $x$ -intercepts are: $x_{i n t} = 1, 4$
Draw a neat sketch

Sketching graphs

Given f ( x ) = x 3 + x 2 - 5 x + 3 :
1. Show that $(x - 1)$ is a factor of $f (x)$ and hence fatorise $f (x)$ fully.
2. Find the coordinates of the intercepts with the axes and the turning points and sketch the graph
Sketch the graph of $f (x) = x^{3} - 4 x^{2} - 11 x + 30$ showing all the relative turning points and intercepts with the axes.
1. Sketch the graph of $f (x) = x^{3} - 9 x^{2} + 24 x - 20$ , showing all intercepts with the axes and turning points.
2. Find the equation of the tangent to $f (x)$ at $x = 4$ .

Local minimum, local maximum and point of inflextion

If the derivative ( $\frac{d y}{d x}$ ) is zero at a point, the gradient of the tangent at that point is zero. It means that a turning point occurs as seen in the previous example.

From the drawing the point (1;0) represents a local minimum and the point (3;4) the local maximum .

A graph has a horizontal point of inflexion where the derivative is zero but the sign of the sign of the gradient does not change. That means the graph always increases or always decreases.

From this drawing, the point (3;1) is a horizontal point of inflexion, because the sign of the derivative does not change from positive to negative.

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Source: OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2

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