Unit 3

Curve Sketching

Lesson 3: Maximum and Minimum Values

Example 4

Give the function

y = f (x)

complete the table below by describing the derivative at the points

a - h

Point	Description of Derivative
$x = a$
$x = b$
$x = c$
$x = d$
$x = e$
$x = f$
$x = g$
$x = h$

State the absolute and local extrema of the function.

Solution

Point	Description of Derivative
$x = a$	undefined
$x = b$	negative
$x = c$	zero
$x = d$	positive
$x = e$	does not exist
$x = f$	zero
$x = g$	zero
$x = h$	negative

Note:

At points

c

f

and

g

, the derivative is zero because the slope of the tangent to the curve is zero (the tangent lines are horizontal).

At point

e

, the derivative does not exist. There is a cusp at point

e

, and as the cusp is approached from the left, the slope of the curve is not the same as when the cusp is approached from the right. As such, the derivative of the function at the cusp does not exist.

Since the function is not defined at

x = a

, there is no absolute maximum.

Since

f (x) \to - \infty

x \to \infty

and

f (x) \to - \infty

x \to - \infty

, there is no absolute minimum.

There are local minima at

x = c

and

x = f

.

There are local maxima at

x = e

and

x = g

Example 5

Verify the graph of the function

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

has a local maximum at

x = 0

Solution

Find the derivative and the critical point(s).

\begin{array}{rcl} f (x) & = & - 3 x^{\frac{2}{3}} \\ f' (x) & = & - 3 (\frac{2}{3}) x^{\frac{2}{3} - 1} \\ = & - 2 x^{- \frac{1}{3}} \\ = & \frac{- 2}{x^{\frac{1}{3}}} \end{array}

The derivative function is not defined (or does not exist – DNE) when the denominator equals zero.

\begin{array}{rcl} x^{\frac{1}{3}} & \neq & 0 \\ x & \neq & 0 \end{array}

Thus,

x = 0

is a critical point.

\begin{array}{rcl} f (x) & = & - 3 x^{\frac{2}{3}} \\ f (0) & = & - 3 {(0)}^{\frac{2}{3}} \\ = & 0 \end{array}

Use the critical point to find the intervals of increase and decrease.

\begin{array}{rclcccrcl} For x < 0, use x & = & - 1 . & For x > 0, use x & = & 1 . \\ f' (x) & = & \frac{- 2}{x^{\frac{1}{3}}} & f' (x) & = & \frac{- 2}{x^{\frac{1}{3}}} \\ f' (- 1) & = & \frac{- 2}{{(- 1)}^{\frac{1}{3}}} & f' (1) & = & \frac{- 2}{{(1)}^{\frac{1}{3}}} \\ = & 2 & = & - 2 \end{array}

Set up a chart to summarize the information found.

Critical Points	$0$
Intervals	$x < 0$	$x > 0$
Sign of $f' (x)$	$+$	$-$
Behaviour of $f (x)$	increasing	decreasing

The function is increasing on the interval

(- \infty, 0)

and decreasing on the interval

(0, \infty)

. Therefore, the point

(0, 0)

is a local maximum, as shown in the graph below.