Question

Find the increasing and decreasing intervlas for the funtion below
f(x)= (x^2-4)^2/3​

Answer 1

Increasing on the interval: (-2, 0) ∪ (2, ∞)

Decreasing on the interval: (-∞, -2) ∪ (0, 2)

Explanation:

A function is increasing when f'(x) > 0

A function is decreasing when f'(x) < 0

Therefore, to find the intervals for which the function is increasing and decreasing, differentiate the given function.

`\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for 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}}`

Given function:

`f(x)=(x^2-4)^{(2)/(3)}`

`\begin{aligned}\textsf{Let} \; u &= (x^2-4) \quad &\implies y&=u^{(2)/(3)}\\\\\implies \frac{\text{d}u}{\text{d}x}&=2x \quad &\implies \frac{\text{d}y}{\text{d}u}&=(2)/(3)u^{-(1)/(3)}=(2)/(3) (x^2-4)^{-(1)/(3)}\end{aligned}`

Therefore:

`\implies \frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}*\frac{\text{d}u}{\text{d}x}`

`\implies \frac{\text{d}y}{\text{d}x}=(2)/(3) (x^2-4)^{-(1)/(3)} * 2x`

`\implies f'(x)=(4)/(3)x (x^2-4)^{-(1)/(3)}`

`\implies f'(x)=\frac{4x}{3 \sqrt[3]{x^2-4}}}`

Stationary points occur when f'(x) = 0:

`\implies \frac{4x}{3 \sqrt[3]{x^2-4}}}=0`

`\implies 4x=0`

`\implies x=0`

Therefore, the function has is a stationary point (turning point) when x = 0.

The derivative is undefined when the denominator equals zero.

The denominator equals zero when x² = 4, so when x = ±2.

The derivative is positive when x > 2 and negative when x < -2.

Therefore, determine the nature of the derivative in the intervals between the stationary point x = 0 and x = ±2.

`\implies f'(-1)=\frac{4(-1)}{3 \sqrt[3]{(-1)^2-4}}}=(\rm negative)/(\rm negative)= \rm positive > 0`

`\implies f'(1)=\frac{4(1)}{3 \sqrt[3]{(1)^2-4}}}=(\rm positive)/(\rm negative)= \rm negative < 0`

Therefore, the function is:

• Increasing on the interval: (-2, 0) ∪ (2, ∞)
• Decreasing on the interval: (-∞, -2) ∪ (0, 2)