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

Question

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

those you have to check

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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