How do you find the local extrema for #f(x)=5x-x^2#?

1 Answer
Mar 8, 2016

Answer:

I found #x=5/2 and y=25/4# as coordinates of a point of maximum for our function.

Explanation:

We can derive our function (representing a Parabola) and set the derivative equal to zero: this will give us the #x# coordinate where the function neither increases nor decreases.
So:
#f'(x)=5-2x#
set equal to zero:
#5-2x=0#
and:
#x=5/2#
which corresponds to a #y# value of:
#f(5/2)=5(5/2)-(5/2)^2=25/2-25/4=25/4#

Graphically:
graph{5x-x^2 [-20.28, 20.28, -10.14, 10.14]}