Answer to Question #308961 in Calculus for Saifi

Question

Find absolute maximum and minimum of the function f(x) = 2x^2- 5 in [-1, 2].

Answer 1

Calculate the derivative:

$f'(x)=4x$

The function can reach an extremum at the boundary points of the segment or in the zeros of the derivative(in this case $x=0$) , calculate:

$f(-1)=-3\\ f(0)=-5\\ f(2)=3$

Answer: $\max=3,\ \min=-5$