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

Find the absolute extrema of $f\left(x\right)=7{x}^{\frac{1}{3}}-{x}^{\frac{7}{3}}$ over the interval [-8,8].
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Annie Levasseur
Step 1
Given function is
$f\left(x\right)=7{x}^{\frac{1}{3}}-{x}^{\frac{7}{3}}$
${f}^{\prime }\left(x\right)=\frac{7}{3}{x}^{\frac{1}{3}-1}-\frac{7}{3}{x}^{\frac{7}{3}-1}$
${f}^{\prime }\left(x\right)=\frac{7}{3{x}^{\frac{2}{3}}}-\frac{7}{3}{x}^{\frac{4}{3}}$
${f}^{\prime }\left(x\right)=\frac{7}{3}\left(\frac{1}{{x}^{\frac{2}{3}}}-{x}^{\frac{4}{3}}\right)=\frac{7}{3}\left(\frac{1-{x}^{\frac{4}{3}}\cdot {x}^{\frac{2}{3}}}{{x}^{\frac{2}{3}}}\right)=\frac{7}{3}\left(\frac{1-{x}^{2}}{{x}^{\frac{2}{3}}}\right)$
Step 2
Now, we solve f'(x)=0 for x
f'(x)=0
$\frac{1-{x}^{2}}{{x}^{\frac{2}{3}}}=0$
$1-{x}^{2}=0$
$x=±1$
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
$f\left(x\right)=7{x}^{\frac{1}{3}}-{x}^{\frac{7}{3}}$
$f\left(-8\right)=7{\left(-8\right)}^{\frac{1}{3}}-{\left(-8\right)}^{\frac{7}{3}}=7\left(-2\right)-{\left(-2\right)}^{7}=-114$
$f\left(-1\right)=7{\left(-1\right)}^{\frac{1}{3}}-{\left(-1\right)}^{\frac{7}{3}}=7\left(-1\right)-{\left(-1\right)}^{7}=-6$
$f\left(0\right)=7{\left(0\right)}^{\frac{1}{3}}-{\left(0\right)}^{\frac{7}{3}}=0$
$f\left(1\right)=7{\left(1\right)}^{\frac{1}{3}}-{\left(1\right)}^{\frac{7}{3}}=7\left(1\right)-{\left(1\right)}^{7}=6$