Walter Clyburn
2021-12-06
Answered

Find the absolute extrema of $f\left(x\right)=7{x}^{\frac{1}{3}}-{x}^{\frac{7}{3}}$ over the interval [-8,8].

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Annie Levasseur

Answered 2021-12-07
Author has **30** answers

Step 1

Given function is

$f\left(x\right)=7{x}^{\frac{1}{3}}-{x}^{\frac{7}{3}}$

$f}^{\prime}\left(x\right)=\frac{7}{3}{x}^{\frac{1}{3}-1}-\frac{7}{3}{x}^{\frac{7}{3}-1$

$f}^{\prime}\left(x\right)=\frac{7}{3{x}^{\frac{2}{3}}}-\frac{7}{3}{x}^{\frac{4}{3}$

${f}^{\prime}\left(x\right)=\frac{7}{3}(\frac{1}{{x}^{\frac{2}{3}}}-{x}^{\frac{4}{3}})=\frac{7}{3}\left(\frac{1-{x}^{\frac{4}{3}}\cdot {x}^{\frac{2}{3}}}{{x}^{\frac{2}{3}}}\right)=\frac{7}{3}\left(\frac{1-{x}^{2}}{{x}^{\frac{2}{3}}}\right)$

Step 2

Now, we solve f'(x)=0 for x

f'(x)=0

$\frac{1-{x}^{2}}{{x}^{\frac{2}{3}}}=0$

$1-{x}^{2}=0$

$x=\pm 1$

Also. f'(x) is not defined for x=0, so all critical numbers of f(x) are -1,0,1

Step 3

Now, we find value of f at the end-points of interval and critical points

$f\left(x\right)=7{x}^{\frac{1}{3}}-{x}^{\frac{7}{3}}$

$f(-8)=7{(-8)}^{\frac{1}{3}}-{(-8)}^{\frac{7}{3}}=7(-2)-{(-2)}^{7}=-114$

$f(-1)=7{(-1)}^{\frac{1}{3}}-{(-1)}^{\frac{7}{3}}=7(-1)-{(-1)}^{7}=-6$

$f\left(0\right)=7{\left(0\right)}^{\frac{1}{3}}-{\left(0\right)}^{\frac{7}{3}}=0$

$f\left(1\right)=7{\left(1\right)}^{\frac{1}{3}}-{\left(1\right)}^{\frac{7}{3}}=7\left(1\right)-{\left(1\right)}^{7}=6$

###### Not exactly what you’re looking for?

Given function is

Step 2

Now, we solve f'(x)=0 for x

f'(x)=0

Also. f'(x) is not defined for x=0, so all critical numbers of f(x) are -1,0,1

Step 3

Now, we find value of f at the end-points of interval and critical points

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How can I progress after this. Every suggestion will be appreciated.

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How can I progress after this. Every suggestion will be appreciated.