Question

Locate the absolute extrema of the function on the closed interval

Answer 1

The absolute extrema is minimum at (-1, 2/9)

Explanation:

Absolute extrema is a logical point that shows whether a the curve function is maximum or minimum.

Forexample a curve in the image attached. A, B and C are points of absolute maxima or absolute maximum. and P and Q are points of absolute minima or minimum.

Remember A, B, C, P, Q are critical points or stationary points.

How do we find absolute extrema?

The find the sign of the second derivative of the function.

From the question;

`{ \sf{g(x) = \sqrt[3]{x} }} \\ \\ { \sf{g(x) = {x}^{ (1)/(3) } }} \\`

Find the first derivative of g(x)

`{ \sf{g {}^(l)(x) = (1)/(3) {x}^{ - (2)/(3) } }} \\`

Find the second derivative;

`{ \sf{g {}^(ll) (x) = ( (1)/(3) * - (2)/(3)) {x}^{( - (2)/(3) - 1) } }} \\ \\ { \sf{g {ll}^((x)) = - (2)/(9) {x}^{ - (5)/(3) } }}`

Then substitute for x as -1 from [-1, 1]

`{ \sf{g {}^(ll)(x) = - (2)/(9) ( - 1) {}^{ - (5)/(3) } }} \\ \\ = ( - 2)/(9) * - 1 \\ \\ = (2)/(9)`

Since the sign of the result is positive, the absolute extrema is minimum