Practice Problems

1. $\mathrm{c} \approx 3,10$, and $13$.

3. (a) $\mathrm{c}=\pi / 2$ (b) $\mathrm{c}=3 \pi / 2,5 \pi / 2,7 \pi / 2,9 \pi / 2$

5. Rolle's Theorem asserts that the velocity $\mathrm{h}^{\prime}(\mathrm{t})$ will equal $0$ at some point between the time the ball is tossed and the time it comes back down. The ball is not moving as fast when it reaches the balcony from below.

7. The function does not violate Rolle's Thm. because the function does not satisfy the hypotheses of the theorem: $\mathrm{f}$ is not differentiable at $0$, a point in the interval $-1 < \mathrm{x} < 1$.

9. No. The velocity is not the same as the rate of change of altitude, since altitude is only one of the components of position. Rolle's Theorem only says there was a time when my altitude was not changing.

11. Since $\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+5, \mathrm{f}^{\prime}(\mathrm{x})=0$ has no real roots. If $\mathrm{f}(\mathrm{x})=0$ for a value of $\mathrm{x}$ other than 2, then by the corollary from Problem 8, we would have an immediate contradiction.

13. (a) $\mathrm{f}(0)=0$, $\mathrm{f}(2)=4$, $\mathrm{f}^{\prime}(\mathrm{c})=2 c$. $\frac{4-0}{2-0}=2 \mathrm{c}$ implies that $\mathrm{c}=1$.
(b) $f(1)=4, f(5)=8$, $f^{\prime}(c)=2 c-5$. $\frac{8-4}{5-1}=2 c-5$ implies that $c=3$.

15. (a) $f(1)=4, f(9)=2$, $f^{\prime}(c)=\frac{-1}{2 \sqrt{c}} \cdot \frac{2-4}{1-9}=\frac{-1}{2 \sqrt{c}}$ implies $\frac{-1}{4}=\frac{-1}{2 \sqrt{c}}$ so $c=4$.
(b) $f(1)=3$, $f(7)=15$, $f^{\prime}(c)=2$. $\frac{15-3}{7-1}=2$ so any $c$ between $1$ and $7$ will do.

17. The hypotheses are not all satisfied since $\mathrm{f}^{\prime}(\mathrm{x})$ does not exist at $\mathrm{x}=0$ which is between $-1$ and $3$.

19. Guilty. All we know is that $\mathrm{f}^{\prime}(\mathrm{c})=17$ at some point, but this does not prove that the motorist "could not have been speeding".

21. $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2}+5 \mathrm{x}+\mathrm{c} . \mathrm{f}(1)=7+\mathrm{c}=10$ when $\mathrm{c}=3$. Therefore, $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2}+5 \mathrm{x}+3$.

23. (a) $\quad \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{Ax}$. We need $\mathrm{A}(1)^{2}+\mathrm{B}=9$ and $2 \mathrm{~A}(1)=4$ so $\mathrm{A}=2$ and $\mathrm{B}=7$ and $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{2} + 7$.
(b) $\mathrm{A}(2)^{2}+\mathrm{B}=3$ and $2 \mathrm{~A}(2)=-2$ so $\mathrm{A}=-1 / 2$ and $\mathrm{B}=5$ and $\mathrm{f}(\mathrm{x})=\frac{-1}{2} \mathrm{x}^{2}+5$.
(c) $\mathrm{A}(0)^{2}+\mathrm{B}=2$ and $\mathbf{2} \mathbf{A}(0)=3$. There is no such $\mathrm{A}$. The point $(0,2)$ is not on the parabola $y=x^{2}+3 x-2$

25. $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{C}$, a family of "parallel" curves for different values of $\mathrm{C}$.

27. $\mathrm{v}(\mathrm{t})=300$. Assuming the rocket left the ground at $\mathrm{t}=0$, we have $\mathrm{y}(1)=300 \mathrm{ft}, \mathrm{y}(2)=600 \mathrm{ft}, \mathrm{y}(5)=1500 \mathrm{ft}$.

29. $\mathrm{f}^{\prime \prime}(\mathrm{x})=6, \mathrm{f}^{\prime}(0)=4, \mathrm{f}(0)=-5 . \mathrm{f}(\mathrm{x})=3 \mathrm{x}^{2}+4 \mathrm{x}-5$.

31. (a) $\mathrm{A}(\mathrm{x})=3 \mathrm{x}$
(b) $A^{\prime}(x)=3$.

33. (a) $\mathrm{A}(\mathrm{x})=\mathrm{x}^2+\mathrm{x}$
(b) $A^{\prime}(x)=2x + 1$.

35. $\mathrm{a}_{1}=5, \mathrm{a}_{2}=\mathrm{a}_{1}+3=5+3=8, \mathrm{a}_{3}=\mathrm{a}_{2}+3=(5+3)+3=11, \mathrm{a}_{4}=\mathrm{a}_{3}+3=(5+3)+3+3=14$.

In general, $a_{n}=5+3(n-1)=2+3 n$.