Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 26

Given that $f(3) = 2$, $f'(3) = \frac{1}{2}$ and $f'(x) > 0$ and $f''(x) < 0$ for all $x$
a.) Sketch the possible graph of $f$
b.) How many solutions does the equation $f(x) = 0$ have? Why?
c.) Is it possible that $f'(2) = \frac{1}{3}$? Why?

a.) The condition tells us that the function has a positive slope and downward concavity at all values of $x$, so the graph might look like this...



b.) There is only one solution for $f(x) = 0 $ since we have a condition of $f'(x) > 0$. When we say $f'(x) > 0$ it means that a s $x$ increases, the value of $f(x)$ will ever increase (positive slope). That means that once the function crosses the $x$-axis, it will never decrease to cross it again.

c.) It is possible that $f'(2) = \frac{1}{3}$ because it satisfies the given condition $f''(x) > 0$