(a) Given y = x² then y’ = 2x and y” = 2. We locate the position of maximum and minimum points by solving y’ = 0 and so such a point exists at x = 0. Evaluating y” at this point we see that y”(0) = 2 which is positive. Using the second-derivative test we conclude that the point is a minimum.

(b) Given y = -t² + t + 1 then y’ = -2t + 1 and y” = -2. Solving y’ = 0 we find t=\frac{1}{2} \text {. Evaluating } y^{\prime \prime} \text { at this point we find } y^{\prime \prime}\left(\frac{1}{2}\right)=-2 which is negative. Using the second-derivative test we conclude that t=\frac{1}{2} is a maximum point.

(c) Given y=\frac{x^3}{3}+\frac{x^2}{2}-2 x+1, \text { then } y^{\prime}=x^2+x-2 \text { and } y^{\prime \prime}=2 x+1 . y^{\prime}=0 at x = 1 and x = -2. At x = 1, y” = 3 which is positive and so the point is a minimum. At x = -2, y” = -3 which is negative and so the point is a maximum.

(d) y^{\prime}=\left\{\begin{array}{cc} -1 & t<0 \\ 1 & t>0 \\ \text { undefined at } t=0 \end{array}\right.

Since y'(0) is undefined, we use the first-derivative test. This was employed in Example 12.1.