question 10\n6 pts\nfind the absolute maximum value of the function ( f(x)=x - sqrt{x} ) on the interval…

question 10\n6 pts\nfind the absolute maximum value of the function ( f(x)=x - sqrt{x} ) on the interval (0,25).\nabsolute max ( f(25)=20 )\nabsolute max ( f(0)=0 )\nabsolute max ( f(25)=0 )\nabsolute max ( f(0)=20 )\nno correct answer choice is given.

question 10\n6 pts\nfind the absolute maximum value of the function ( f(x)=x - sqrt{x} ) on the interval (0,25).\nabsolute max ( f(25)=20 )\nabsolute max ( f(0)=0 )\nabsolute max ( f(25)=0 )\nabsolute max ( f(0)=20 )\nno correct answer choice is given.

Answer

Explanation:

Step1: Find the derivative of the function

The function is (f(x)=x - \sqrt{x}=x - x^{\frac{1}{2}}). Using the power rule ((x^n)^\prime=nx^{n - 1}), the derivative (f^\prime(x)=1-\frac{1}{2}x^{-\frac{1}{2}}=1-\frac{1}{2\sqrt{x}}).

Step2: Find the critical points

Set (f^\prime(x) = 0), then (1-\frac{1}{2\sqrt{x}}=0). [ \begin{align*} \frac{1}{2\sqrt{x}}&=1\ 2\sqrt{x}&=1\ \sqrt{x}&=\frac{1}{2}\ x&=\frac{1}{4} \end{align*} ] Also, check the endpoints of the interval (x = 0) and (x = 25).

Step3: Evaluate the function at critical points and endpoints

  • When (x = 0), (f(0)=0-\sqrt{0}=0).
  • When (x=\frac{1}{4}), (f(\frac{1}{4})=\frac{1}{4}-\sqrt{\frac{1}{4}}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}).
  • When (x = 25), (f(25)=25-\sqrt{25}=25 - 5=20).

Answer:

Absolute max (f(25)=20)