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

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Answer 1 · 2015-11-10T14:18:11

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 $f(-4)$ .

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