Question

determine the location and value of the absolute extreme values of f on the given interval, if they exist
f(x)=4 x^{\frac{3}{4}}-x \text { 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 27 at ( x=81 )
(use a comma to separate answers as needed)
b. there is no absolute maximum of f on the given interval
what is/are the absolute minimum/minima 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 minimum/minima is/are ( quad ) at ( x=quad )
(use a comma to separate answers as needed)
b. there is no absolute minimum of f on the given interval

Explanation:

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

Using the power rule \( (x^n)^\prime=nx^{n - 1} \), for \( f(x)=4x^{\frac{3}{4}}-x \), we have \( f^\prime(x)=4\times\frac{3}{4}x^{\frac{3}{4}-1}-1 = 3x^{-\frac{1}{4}}-1=\frac{3}{\sqrt[4]{x}}-1 \)

Step2: Find the critical points

Set \( f^\prime(x) = 0 \), then \( \frac{3}{\sqrt[4]{x}}-1 = 0 \).

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

Answer:

For the absolute maximum:
A. The absolute maximum/maxima is/are \( 27 \) at \( x = 81 \)
For the absolute minimum:
A. The absolute minimum/minima is/are \( 0 \) at \( x = 0,256 \)