.
Back to the class
Section 3.2 #55: The function
$$f(x) = \left\{ \begin{array}{ll}
0, & x=0 \\
1-x, & 0 < x \leq 1
\end{array} \right.$$
is differentiable on $(0,1)$ and satisfies $f(0)=f(1)$. However, its derivative is never zero on $(0,1)$. Does this contradict Rolle's theorem? Explain.
Solution: This does not contradict Rolle's theorem, because this function is not continuous at $x=0$ (draw the graph!).
Section 3.2 #70: Find a function that has the derivative $f'(x)=4$ and whose graph passes through point $(0,1)$. Explain your reasoning.
Solution: The function with these properties is $f(x)=4x+1$. We pick it because $\dfrac{\mathrm{d}}{\mathrm{d}x}[4x]=4$ satisfying the condition $f'(x)=4$ and because $f(x)=4x$ is not good enough (it does not pass through $(0,1)$) and so we add the $1$ to make it satisfy that condition. Notice that adding a constant does not change the value of its derivative because the derivative of a constant is zero!