Question

What is the absolute extrema of the function on the closed interval f (x)=x^2+2x-4, [-1,1]

Answer 1

$f(x)=x^2+2x-4\\ f'(x)=2x+2\\\\ 2x+2=0\\ 2x=-2\ x=-1\\\\ f(-1)=(-1)^2+2\cdot(-1)-4=1-2-4=-5\\ f(1)=1^2+2\cdot1-4=1+2-4=-1\\\\ f_{min}=-5\\ f_{max}=-1$