Question

Find the value of the derivative (if it exists) at
each indicated extremum

Answer 1

The derivative does not exist at the extremum (-2, 0).

Explanation:

Given function:

`f(x)=(x+2)^{(2)/(3)}`

To differentiate the given function, use the chain rule and the power rule of differentiation.

`\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule of Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}*\frac{\text{d}u}{\text{d}x}$\\\end{minipage}}`


`\boxed{\begin{minipage}{5.4 cm}\underline{Power Rule of Differentiation}\\\\If $y=x^n$, then $\frac{\text{d}y}{\text{d}x}=nx^(n-1)$\\\end{minipage}}`

`\begin{aligned}\textsf{Let}\;u &= x+2& \implies f(u) &= u^{(2)/(3)}\\\\\implies \frac{\text{d}u}{\text{d}{x}}&=1 &\implies \frac{\text{d}y}{\text{d}u}&=(2)/(3)u^{((2)/(3)-1)}=(2)/(3)u^{-(1)/(3)}\end{aligned}`

Apply the chain rule:

`\implies f'(x) = \frac{\text{d}y}{\text{d}{u}} \cdot \frac{\text{d}u}{\text{d}{x}}`

`\implies f'(x) = (2)/(3)u^{-(1)/(3)} \cdot1`

`\implies f'(x) = (2)/(3)u^{-(1)/(3)}`

Substitute back in u = x + 2:

`\implies f'(x) = (2)/(3)(x+2)^{-(1)/(3)}`

`\implies f'(x) = \frac{2}{3(x+2)^{(1)/(3)}}`

An extremum is a point where a function has a maximum or minimum value. From inspection of the given graph, the minimum point of the function is (-2, 0).

To determine the value of the derivative at (-2, 0), substitute x = -2 into the differentiated function.

`\begin{aligned}\implies f'(-2) &= \frac{2}{3(-2+2)^{(1)/(3)}}\\\\ &= \frac{2}{3(0)^{(1)/(3)}}\\\\&=(2)/(0) \;\;\;\leftarrow \textsf{unde\:\!fined}\end{aligned}`

As the denominator of the differentiated function at x = -2 is zero, the value of the derivative at (-2, 0) is undefined. Therefore, the derivative does not exist at the extremum (-2, 0).