Question

determine the location and value of the absolute extreme values of f on the given interval, if they exist

$f(x)=4x^{\frac{3}{4}}-x$ on $0,256$

what is/are the absolute maximum/maxima of f on the given interval? select the correct choice below and, if necessary, fill in the answer boxes to complete your choice

a. the absolute maximum/maxima is/are at $x=$
(use a comma to separate answers as needed.)
b. there is no absolute maximum of f on the given interval

Explanation:

Step1: Find the derivative of \(f(x)\)

Use the power rule \((x^n)^\prime=nx^{n - 1}\).
For \(f(x)=4x^{\frac{3}{4}}-x\), \(f^\prime(x)=4\times\frac{3}{4}x^{\frac{3}{4}-1}-1 = 3x^{-\frac{1}{4}}-1=\frac{3}{\sqrt[4]{x}}-1\)

Step2: Find the critical points

Set \(f^\prime(x) = 0\), \(\frac{3}{\sqrt[4]{x}}-1=0\)
\(\frac{3}{\sqrt[4]{x}}=1\), then \(\sqrt[4]{x}=3\), so \(x = 81\)
Also, check the endpoints \(x = 0\) and \(x = 256\)

Step3: Evaluate \(f(x)\) at critical points and endpoints

• When \(x = 0\), \(f(0)=4\times0^{\frac{3}{4}}-0=0\)
• When \(x = 81\), \(f(81)=4\times81^{\frac{3}{4}}-81=4\times(3^4)^{\frac{3}{4}}-81=4\times3^3 - 81=4\times27-81=108 - 81 = 27\)
• When \(x = 256\), \(f(256)=4\times256^{\frac{3}{4}}-256=4\times(4^4)^{\frac{3}{4}}-256=4\times4^3-256=4\times64 - 256=256 - 256=0\)

Answer:

A. The absolute maximum/maxima is/are \(27\) at \(x = 81\)