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

determine the location and value of the absolute extreme values of f on the given interval, if they exist\n\n$f(x)=4x^{\\frac{3}{4}}-x$ on $0,256$\n\nwhat 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\n\na. the absolute maximum/maxima is/are at $x=$\n(use a comma to separate answers as needed.)\nb. there is no absolute maximum of f on the given interval

determine the location and value of the absolute extreme values of f on the given interval, if they exist\n\n$f(x)=4x^{\\frac{3}{4}}-x$ on $0,256$\n\nwhat 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\n\na. the absolute maximum/maxima is/are at $x=$\n(use a comma to separate answers as needed.)\nb. there is no absolute maximum of f on the given interval

Answer

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)