# Determine the intervals on which the function is concave up or concave down. f(x)=x(x-8\sqrt{x})...

## Question:

Determine the intervals on which the function is concave up or concave down.

{eq}f(x)=x(x-8\sqrt{x}) \quad |x > 0| {/eq}

## Concavity:

Recall that

1) If {eq}f''(x)>0 {/eq} on an interval {eq}I {/eq}, then {eq}f(x) {/eq} is concave up on {eq}I. {/eq}

2) If {eq}f''(x)<0 {/eq} on an interval {eq}I, {/eq} then {eq}f(x) {/eq} is concave down on {eq}I. {/eq}

Suppose {eq}f(x) {/eq} is defined at {eq}x=a {/eq} and {eq}f(x) {/eq} changes concavity at {eq}x=a, {/eq} then {eq}f(x) {/eq} has an inflection point at {eq}x=a. {/eq}

Let {eq}f(x)=x(x-8\sqrt{x})=x^2-8x^{3/2},\, x>0 {/eq}. We then have

{eq}f'(x)=2x-12x^{1/2}\\ f''(x)=2-6x^{-1/2} {/eq}

Setting this equal to 0, we have

{eq}2-6x^{-1/2}=0\\ \frac{6}{x^{1/2}}=2\\ x^{1/2}=3\\ x=9 {/eq}

Next we note

On the interval {eq}[0,9) {/eq}, we have {eq}f''(x)>0 {/eq} (use the test point {eq}x=1 {/eq} if needed).

On the interval {eq}(9,\infty) {/eq}, we have {eq}f''(x)<0 {/eq} (use the test point {eq}x=10 {/eq} if needed).

We conclude that {eq}f(x) {/eq} is concave up on {eq}(0,9) {/eq} and concave down on {eq}(9,\infty) {/eq}. 