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interval -2,3 not -23

use extreme value theorem

continuous functions on a compact set attain an absolute max and min

Oops, I meant "Similarly, the second derivative is positive at -1 and 2......"

you don't need the 2nd derivative

i have a whole bunch of calculus that i need help with

m(x)=3x^4−4x^3−12x^2
for x =2 y = 48-32-48 = -32
for x = -1 y =3-4+12=11

What is the maximum and min value of f(x)=t2/t2+3 on the interval [−1,1] ?

I assume that's
\[f(x) = \frac{t^2}{t^2 + 3} \]
right?

yes

do you have time for a few more ?