What are the extrema of #f(x)=x^2-192x+8 # on #x in[-4,9]#?

Question

What are the extrema of  #f(x)=x^2-192x+8 # on #x in[-4,9]#?

Isaiah Allen · Accepted Answer

The minimum is #f(9)#, and the maximum is #f(-4)#.

#f'(x) = 2x-192#, so there are no critical numbers for #f# in the interval chosen. Therefore, The minimum and maximum occur at the endpoints. #f(-4) = 16+192(4)+8# is clearly a positive number and #f(9)=81-192(9)+4# is clearly negative. So, the minimum is #f(9)#, and the maximum is #f(-4)#.

Isabella Ackerman · Answer

undefined To find the extrema of $ f(x) = x^2 - 192x + 8 $ on $ x $ in the interval $ [-4, 9] $:

1. Find the critical points by taking the derivative of $ f(x) $ and setting it equal to zero.
2. Evaluate $ f(x) $ at the critical points and at the endpoints of the interval.
3. Identify the maximum and minimum values among these values.

First, find the derivative of $ f(x) $:

$$ f'(x) = 2x - 192 $$

Set $ f'(x) $ equal to zero to find the critical points:

$$ 2x - 192 = 0 $$
$$ x = 96 $$

Evaluate $ f(x) $ at the critical point and the endpoints of the interval $ [-4, 9] $:

At $ x = -4 $: $ f(-4) = (-4)^2 - 192(-4) + 8 = 360 $

At $ x = 9 $: $ f(9) = (9)^2 - 192(9) + 8 = -1459 $

At $ x = 96 $: $ f(96) = (96)^2 - 192(96) + 8 = -9016 $

The critical point $ x = 96 $ lies outside the interval $ [-4, 9] $, so we only consider $ x = -4 $ and $ x = 9 $ within the interval.

The minimum value occurs at $ x = 9 $, where $ f(x) = -1459 $.

The maximum value occurs at $ x = -4 $, where $ f(x) = 360 $.