Question

Absolute max/min! f(x)=1-7x^2, on the interval -3 less than or equal to x less than or equal to 2.

Answer 1

max is at x = 0 (so max value is 1)
check the endpoints for the min

Answer 2

that is, check
\[f(-3)\] and
\[f(2)\] whatever is smallest is smallest. i will do doubt be
\[f(-3)\] since -3 is further from 0 than 2 is

Answer 3

well I had the max no problem, but I can't figure out the min value

Answer 4

those you have to check

Answer 5

Yeah, I can't seem to solve it. Thank you though

Answer 6

\[f(-3)=1-7\times 9=-62\] is the min

Answer 7

because
\[f(2)=1-7\times 4=-27\] and
\[-62<-27\] so
\[-62\] is the mininum value on that interval

Answer 8

Thank you so much! :D you are a life saver