Section 3.7 « Difference Equations to Differential Equations

Section 3.7

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1. Let $g(x) = \cos(x) - x$ . Then $g(0) = 1$ and $g(1) = \cos(1) - 1 < 0$ (since $\cos(x) < 1$ ). Hence, by the Intermediate Value Theorem, there is at least one point $c$ in the interval $(0, 1)$ such that $g(c) = 0$ . Now if there were two points, say, $c_1$ and $c_2$ , in $(0, 1)$ such that $g(c_1) = 0$ and $g(c_2) = 0$ , then, by Rolle’s Theorem, there would be a point $d$ between $c_1$ and $c_2$ such that $g'(d) = 0$ . But this cannot happen since $g'(x) = -\sin(x) - 1 < 0$ for all $x$ in $(0, 1)$ . Hence there is exactly one point $c$ in $(0, 1)$ such that $g(c) = 0$ . That is, there is exactly one solution to the equation $\cos(x) = x$ in the interval $(0, 1)$ .

3. By the Mean Value Theorem, for any two points $u$ and $v$ in $[a, b]$ , there exists a point $c$ in $(a, b)$ such that

$\displaystyle{f'(c) = \frac{f(v) - f(u)}{v - u}}$ .

Hence $f(v) - f(u) = f'(c)(v - u)$ , and so

$|f(v) - f(u)| = |f'(c)||v - u| \le M|v - u|$ .

5. (a) $f$ is increasing on $(0, \infty)$ and decreasing on $(-\infty, 0)$ .

(e) $f$ is increasing on $(-1, 1)$ and decreasing on both $(-\infty, -1)$ and $(1, \infty)$

(g) $y$ is decreasing on $(-\infty, -2)$ , $(-2, 2)$ , and $(2, \infty)$ . There are no intervals on which $y$ is increasing.

7. (a) Suppose $u$ and $v$ are in $[a, b]$ with $u < v$ . Then, by the Mean Value Theorem, there exists a point $c$ in $(a, b)$ such that

$\displaystyle{f'(c) = \frac{f(v) - f(u)}{v - u}}$ .

By assumption, $f'(c) > 0$ , and so

$f(v) - f(u) = f'(c)(v - u) > 0$ .

Hence $f(v) > f(u)$ ; thus $f$ is increasing on $[a, b]$ .

9. Let

$f(x) = 1 + \dfrac{1}{2}x - \sqrt{1 + x}$ .

Then

$f'(x) = \dfrac{1}{2} - \dfrac{1}{2\sqrt{1 + x}}$ .

Hence $f'(x) > 0$ for all $x > 0$ , and so $f$ is, by Problem 7, increasing on $[0, \infty)$ . In particular, for any $x > 0$ , $f(0) < f(x)$ . Since $f(0) = 0$ , it follows that

$1 + \dfrac{1}{2}x - \sqrt{1 + x} > 0$

for all $x > 0$ .

11. (a) $F(x) = x^2 + k$ is an antiderivative of $f$ for any constant $k$ .

(e) $\displaystyle{H(x) = \frac{1}{3}x^3 - \frac{3}{2}x^2 + k}$ is an antiderivative of $h$ for any constant $k$ .

13. If $G$ is an antiderivative of $g$ , then $G(x) = -\frac{1}{2}\cos(2x) + k$ for some constant $k$ .

15. $k = -1$

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Difference Equations to Differential Equations

Section 3.7

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