What are the extrema of #f(x) = 64-x^2# on the interval #[-8,0]#?

Question

Paisley Johnson · Accepted Answer

Find the critical values on the interval (when #f'(c)=0# or does not exist). #f(x)=64-x^2# #f'(x)=-2x# Set #f'(x)=0#. #-2x=0# #x=0# And #f'(x)# is always defined. To find the extrema, plug in the endpoints and the critical values. Notice that #0# fits both of these criteria. #f(-8)=0larr"absolute minimum"# #f(0)=64larr"absolute maximum"# graph{64-x^2 [-8, 0, -2, 66]}

Samantha Adkison · Answer

undefined The function $ f(x) = 64 - x^2 $ has extrema on the interval $[-8, 0]$. To find these extrema, we first need to find the critical points by taking the derivative of the function and setting it equal to zero.

$ f'(x) = -2x $

Setting $ f'(x) $ equal to zero, we get:

$ -2x = 0 $

Solving for $ x $, we find the critical point at $ x = 0 $.

To determine if this critical point is a maximum or minimum, we can use the second derivative test.

$ f''(x) = -2 $

Since the second derivative is negative, the critical point $ x = 0 $ corresponds to a maximum.

Now we need to check the endpoints of the interval, $ x = -8 $ and $ x = 0 $, to see if they are also extrema.

$ f(-8) = 64 - (-8)^2 = 64 - 64 = 0 $

$ f(0) = 64 - 0^2 = 64 $

So, the function has an extremum at $ x = -8 $ where it reaches a minimum of $ 0 $, and at $ x = 0 $ where it reaches a maximum of $ 64 $.