Practice Problems

1. Discontinuous at $\mathrm{1}$, $\mathrm{3}$, and $\mathrm{4}$.

3. (a) Discontinuous at $\mathrm{x}=3$. Fails condition (i) there.

(b) Discontinuous at $\mathrm{x}=2$. Fails condition (i) there.

(c) Discontinuous where $\cos (x)$ is negative, (e.g., at $x=\pi$). Fails condition (i) there.

(d) Discontinuous where $\mathrm{x}^{2}$ is an integer (e.g., at $\mathrm{x}=1$ or $\sqrt{2}$). Fails condition (ii) there.

(e) Discontinuous where $\sin (x)=0$ (e.g., at $x=0, \pm \pi, \pm 2 \pi, \ldots)$. Fails condition (i) there.

(f) Discontinuous at $\mathrm{x}=0$. Fails condition (i) there.

(g) Discontinuous at $\mathrm{x}=0$. Fails condition (i) there.

(h) Discontinuous at $\mathrm{x}=3$. Fails condition (i) there.

(i) Discontinuous at $\mathrm{x}=\pi / 2$. Fails condition (i) there.

5. (a) $\mathrm{f}(\mathrm{x})=0$ for at least 3 values of $\mathrm{x}, 0 \leq \mathrm{x} \leq 5$.

(b) $\mathrm{1}$

(c) $\mathrm{3}$

(d) $\mathrm{3}$

(e) Yes. It does not have to happen, but it is possible.

7. (a) $\mathrm{f}(0)=0, \mathrm{f}(3)=9$ and $0 \leq 2 \leq 9. \mathrm{c}=\sqrt{2} \approx 1.414$

(b) $f(-1)=1, f(2)=4$ and $1 \leq 3 \leq 4. c=\sqrt{3} \approx 1.732$

(c) $\mathrm{f}(0)=0, \mathrm{f}(\pi / 2)=1$ and $0 \leq 1 / 2 \leq 1. \mathrm{c}=($ inverse sine of $1 / 2) \approx 0.524$

(d) $f(0)=0, f(1)=1$ and $0 \leq 1 / 3 \leq 1. c=1 / 3$

(e) $f(2)=2, f(5)=20$ and $2 \leq 4 \leq 20. c=(1+\sqrt{17}) / 2 \approx 2.561$.

(f) $f(1)=0, f(10) \approx 2.30$ and $0 \leq 2 \leq 2.30. c=$ (inverse of $\ln (2))=e^{2} \approx 7.389$.

9. Neither student is correct. The bisection algorithm converges to the root labeled $\mathrm{C}$.

11. (a) D

(b) $\mathrm{D}$

(c) hits B

13. $[-0.9375,-0.875], \approx-0.879$

$[1.3125,1.375], \approx 1.347$

$[2.5,2.5625], \approx 2.532$

15. $[2.3125,2.375], \approx 2.32$.

17. $[-0.375,-0.3125], \approx-0.32$.

19. See the three graphs in Fig. 1.3P19.

21. (a) $\mathrm{A}(2.1)-\mathrm{A}(2)$ is the area of the region bounded below by the $\mathrm{x}$-axis, above by the graph of $\mathrm{f}$, on the left by the vertical line $\mathrm{x}=2$, and on the right by the vertical line $\mathrm{x}=2.1$.

$\frac{\mathrm{A}(2.1)-\mathrm{A}(2)}{0.1} \approx \mathrm{f}(2) \text { or } \mathrm{f}(2.1) \text { so } \frac{\mathrm{A}(2.1)-\mathrm{A}(2)}{0.1} \approx 1$

(b) $\mathrm{A}(4.1)-\mathrm{A}(4)$ is the area of the region bounded below by the $\mathrm{x}$-axis, above by the graph of $\mathrm{f}$, on the left by the vertical line $x=4$, and on the right by the vertical line $x=4.1 . \frac{A(4.1)-A(4)}{0.1} \approx f(4) \approx 2$.

23. (a) Yes. You supply the justification.

(b) Yes

(c) Try it.