Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises - Page 303: 80

(a) $f+g$ is concave upward on $I$ (b) $g(x)$ is concave upward on $I$

(a) $f'' \gt 0$ and $g'' \gt 0$ on $I$ because they are both concave upward on $I$ $(f+g)' = f'+g'$ $(f+g)'' = (f'+g')' = f''+g'' \gt 0$ on $I$ Therefore, $f+g$ is concave upward on $I$ (b) $f \gt 0$ on $I$ since $f$ is positive on $I$ $f'' \gt 0$ on $I$ since $f$ is concave upward on $I$ $g(x) = [f(x)]^2$ $g'(x) = 2[f(x)]~f'(x)$ $g''(x) = 2f'(x)~f'(x)+2f(x)~f''(x)$ $g''(x) = 2[f'(x)]^2+2f(x)~f''(x) \gt 0$ on $I$ Therefore, $g(x)$ is concave upward on $I$