Find the absolute max value and min value of this function

Question

Find the absolute max value and min value of this function

f(x)= x/x^2+1 on the interval [0,2]

Answer 1

Graph this function.

f'(x) = (1-x²)/(x² + 1)²

f'(1)=0 and f(1)=1/2, the maximum value.

The minimum on the interval is 0 at x=0.

Answer 2

test endpoints f(0)=0, f(2)=2/5

find critical points. take derivative of f(x) using quotient rule.

f'(x)=(x-1)^2/(x^2+1)^2 after simplifying.

f'(x)=0 implies x=1

f(1)=1/2

given f(0)=0, f(1)=1/2, f(2)=2/5 which will be absolute max and absolute min