Question

Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the function to two decimal places. maximum minimum (b) Use calculus to find the exact maximum and minimum values. maximum minimum

Answer 1

ANSWER

See below

EXPLANATION

Part a)

The given function is

$f(x) = {x}^{5} - {x}^{3} + 6$

From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

b) Using calculus, we find the first derivative of the given function.

$f'(x) = 5 {x}^{4} - 3 {x}^{2}$

At turning point f'(x)=0.

$5 {x}^{4} - 3 {x}^{2} = 0$

This implies that,

${x}^{2} (5 {x}^{2} - 3) = 0$

${x}^{2} = 0 \: or \: 5 {x}^{2} - 3 = 0$

$x = - \frac{ \sqrt{15} }{5} \: or \: x = 0 \: \: or \: x =\frac{ \sqrt{15} }{5}$

We plug this values into the original function to obtain the y-values of the turning points

$( - \frac{ \sqrt{15} }{5} , \frac{1}{125} ( 6 \sqrt{15} +750)) \:and \: (0, - 6) \: and\: ( \frac{ \sqrt{15} }{5} , \frac{1}{125} ( - 6 \sqrt{15} +750))$

We now use the second derivative test to determine the absolute maximum minimum on the interval [-1,1]

$f''(x) = 20 {x}^{3} - 6x$

$f''( - \frac{ \sqrt{15} }{5} ) \: < \: 0$

Hence

$( - \frac{ \sqrt{15} }{5} , \frac{1}{125} ( 6 \sqrt{15} + 750))$

is a maximum point.

$f''( \frac{ \sqrt{15} }{5} ) \: > \: 0$

Hence

$( \frac{ \sqrt{15} }{5} , \frac{1}{125} (- 6 \sqrt{15} + 750))$

is a minimum point.

$f''(0) \: =\: 0$

Hence (0,-6) is a point of inflexion