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

determine the location and value of the absolute extreme values of f on the given interval, if they exist\n f(x)=4 x^{\frac{3}{4}}-x \text { on }0,256 \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.\na. the absolute maximum/maxima is/are 27 at ( x=81 )\n(use a comma to separate answers as needed)\nb. there is no absolute maximum of f on the given interval\nwhat 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.\na. the absolute minimum/minima is/are ( quad ) at ( x=quad )\n(use a comma to separate answers as needed)\nb. there is no absolute minimum 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 f(x)=4 x^{\frac{3}{4}}-x \text { on }0,256 \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.\na. the absolute maximum/maxima is/are 27 at ( x=81 )\n(use a comma to separate answers as needed)\nb. there is no absolute maximum of f on the given interval\nwhat 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.\na. the absolute minimum/minima is/are ( quad ) at ( x=quad )\n(use a comma to separate answers as needed)\nb. there is no absolute minimum of f on the given interval

Answer

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 ). [ \begin{align*} \frac{3}{\sqrt[4]{x}}&=1\ \sqrt[4]{x}&=3\ x&=81 \end{align*} ] 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 )