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**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!