(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}.  Evaluating y′′ 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,  then y′ = x² + x − 2 and y′′ = 2x + 1. y′ = 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.