What are the extrema of #f(x)=f(x)= -x^2+8x+7#?

Question

What are the extrema of  #f(x)=f(x)= -x^2+8x+7#?

Charles Anctil · Accepted Answer

it has local maximum at (4;23)

At the turning point f'(x) = 0. Thus -2x +8 = 0. So x= 4 and y = f(4) at extremum (4;23). The shape of the graph (an upside down "cup" indicates that this is a maximum point)

Samantha Adkison · Answer

undefined To find the extrema of the function $ f(x) = -x^2 + 8x + 7 $, you first need to find its critical points by taking the derivative of the function and setting it equal to zero. Then, you can determine whether those critical points correspond to a maximum or minimum by checking the sign of the second derivative.

1. Find the derivative of $ f(x) $: $ f'(x) = -2x + 8 $.
2. Set $ f'(x) $ equal to zero and solve for $ x $:
   $$ -2x + 8 = 0 $$
   $$ -2x = -8 $$
   $$ x = 4 $$
3. Check the sign of the second derivative at $ x = 4 $ to determine if it's a maximum or minimum.
4. Find the second derivative of $ f(x) $: $ f''(x) = -2 $.
5. Since $ f''(4) = -2 $ (negative), the function has a maximum at $ x = 4 $.
6. To find the maximum value, substitute $ x = 4 $ back into the original function:
   $$ f(4) = -4^2 + 8(4) + 7 = 9 $$
   So, the maximum value is $ f(4) = 9 $.

Therefore, the function $ f(x) = -x^2 + 8x + 7 $ has a maximum at $ x = 4 $ with a maximum value of 9.