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

Question

Emma Aldridge · Accepted Answer

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

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]}

Charles Anctil · Answer

undefined To find the local extrema for the function $ f(x) = 5x - x^2 $, we first need to find the critical points by taking the derivative of the function and setting it equal to zero. Then, we can determine whether these critical points correspond to local maxima or minima by using the second derivative test or by analyzing the behavior of the function around these points.

1. Find the derivative of the function:
$$ f'(x) = 5 - 2x $$

2. Set the derivative equal to zero and solve for $ x $ to find critical points:
$$ 5 - 2x = 0 $$
$$ 2x = 5 $$
$$ x = \frac{5}{2} $$

3. Determine the nature of the critical point:
- We can use the second derivative test to determine whether $ x = \frac{5}{2} $ corresponds to a local maximum or local minimum.
- Take the second derivative of the function:
$$ f''(x) = -2 $$

- Since the second derivative is negative for all values of $ x $, the critical point $ x = \frac{5}{2} $ corresponds to a local maximum.

Therefore, the local maximum of the function $ f(x) = 5x - x^2 $ occurs at $ x = \frac{5}{2} $.