Answer
(b). False
(a). True
Work Step by Step
(a) True. Let $x_{1}$ be in the interval; for a function to be concave up, it means that the second derivative is positive. So we get:
\[
(g+f)^{\prime \prime}\left(x_{1}\right)=g^{\prime \prime}\left(x_{1}\right)+f^{\prime \prime}\left(x_{1}\right)>0+0=0
\]
So $g+f$ is concave up.
(b). False Let $g(x)=\frac{1}{x^{2}},f(x)=x^{2} $. The two functions are concave up in the interval $(0, \infty),$ but for the function $1=g \cdot f, 0=f^{\prime }(x), 0= f^{\prime \prime}(x)$ everywhere. So it cannot be concave up.
