Here's the question you clicked on:
soccergal12
find the absolute extrema of the following function on the given interval: f(x) = x^2 / (x^2 + 3) , [-1,1]
I guess you have to use calculus and are not allowed to use a calculator?
yeah, we aren't allowed to use a calculator
\[ f'(x)=\frac{6 x}{\left(x^2+3\right)^2}\\ f'(x) >0 \text { if} x >0\\ f'(x) <0 \text { if} x<0\\ f'(x) =0 \text { if} x=0\\ \]
Do you know how to use the 1st derivative test?
Hence we have to examine f(1)=1/4 f(-1)=1/4 f(0)=0 Max 1/4 and Min=0