How do you find the exact relative maximum and minimum of the polynomial function of #f(x)=4x-x^2#?

Question

Emma Aldridge · Accepted Answer

At #x=2#

#dy/dx=0# and #(d^2y)/(dx^2)<0#

Hence the function has a maximum at #x=2# #y=4x-x^2# #dy/dx=4-2x# #dy/dx=>4-2x=0# #x=(-4)/(-2)=2# #(d^2y)/(dx^2)=-2# At #x=2# #dy/dx=0# and #(d^2y)/(dx^2)<0# Hence the function has a maximum at #x=2# graph{4x-x^2 [-10, 10, -5, 5]}

Adam Adkins · Answer

undefined To find the exact relative maximum and minimum of the polynomial function $ f(x) = 4x - x^2 $:

1. Find the critical points by setting the derivative equal to zero and solving for $ x $.
2. Determine the nature of each critical point by evaluating the second derivative at each point.
3. Confirm whether the critical points are relative maximum, minimum, or points of inflection.

Let's start with finding the derivative of $ f(x) $:
$ f'(x) = 4 - 2x $.

Setting $ f'(x) $ equal to zero:
$ 4 - 2x = 0 $,
$ 2x = 4 $,
$ x = 2 $.

Now, find the second derivative of $ f(x) $:
$ f''(x) = -2 $.

Since $ f''(x) $ is constant and negative, the critical point at $ x = 2 $ is a relative maximum.

Therefore, the exact relative maximum of $ f(x) $ is at $ x = 2 $, and the maximum value is $ f(2) = 4 $.