What are the global and local extrema of #f(x)=4x-x^2 # ?

1 Answer
Feb 12, 2016

Answer:

#x=2# is both a global and local maximum.

Explanation:

Assume that the domain of #f# is all real numbers.

Complete the square

#f(x) = -(x-2)^2-4#

As anything squared cannot be less than zero, #f(2)=-4# is the local maximum as well as the global maximum.

To show using calculus, take the first derivative of #f#.

#f'(x) = 4-2x#

Solve for #f'(x)=0#.

#4-2x=0#

#x=2#

Perform the second derivative test.

#f''(x) = -2#

#f''(2) = -2 < 0#

This means that the gradient is decreasing at #x=2#.

This means that the gradient changes from positive to zero to negative in a small neighborhood around #x=2#.

This means that #x=2# is a local maximum.

There are no other local extrenum and #f# is continuous everywhere.

#x=2# is also a global maximum.