Question

locate the absolute extrema of the function on the closed interval: g(t)= (t^2)/(t^2+3), [-1,1]

Answer 1

Find the derivative:\[g'(t) = \frac{2t\left(t^2+3\right)-2t\left(t^2\right)}{\left(t^2+3\right)^2} = \frac{6t}{\left(t^2+3\right)^2}.\]
Find the points where the derivative is zero:\[g'(t) = \frac{6t}{\left(t^2+3\right)^2} = 0 \Rightarrow 6t = 0 \Rightarrow t = 0\]
Compare the end points of the domain and the points where the derivate is zero:\[g(-1) = \frac{1}{4}\]\[g(0) = 0\]\[g(1) = \frac{1}{4}\]These are the only points that matter, so the minimum of this points is the absolute minimum and the maximum the absolute maximum. Therefore, the absolute minimum is at t = 0 and the absolute maximum at \[t = \pm 1.\]