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 — de2de:chapter-3 [2012/06/25 11:25] (current)dcs created 2012/06/25 11:25 dcs created 2012/06/25 11:25 dcs created Line 1: Line 1: + ====Differential Equations to Difference Equations==== + ===Answers for selected problems=== + + **Chapter 3** + + ---- + + **Section 3.1** + + 1. (b) $m = 1$ + + 2. (a) $o(h)$ + + (b) $O(h)$ + + %%(c)%% $o(h)$ + + (d) $o(h)$ + + (e) $o(h)$ + + (f) Neither ​ + + + 3. (b) $R_T$ is $O(h)$, but not $o(h)$. + + %%(c)%% $R_S$ is both $o(h)$ and $O(h)$. + + (d) $S$ is the best affine approximation to $f$ at $9$. + + (e) $\sqrt{10} \approx S(10) = \displaystyle{\frac{19}{6}} = 3.167$ to three decimal places, whereas $\sqrt{10}$ is exactly $3.162$ to three decimal places. + + + 5. $T(x) = 4(x - 1) + 2$, $f'(1) = 4$ + + 7. $T(x) = x - 1$, $f'(0) = 1$ + + ---- + + **Section 3.2** + + 1. (a) $f'(2) = 4$ + + %%(c)%% $g'(2) = -\displaystyle{\frac{1}{4}}$ + + (e) $f'(1) = -\displaystyle{\frac{1}{2}}$ + + + 2. (a) Best affine approximation:​ $T(x) = 4(x - 2) + 5$ + Equation of tangent line: $y = 4(x - 2) + 5$ + + %%(c)%% Best affine approximation:​ $\displaystyle{T(x) = -\frac{1}{4}(x - 2) + \frac{1}{4}}$ + Equation of tangent line: $\displaystyle{y = -\frac{1}{4}(x - 2) + \frac{1}{4}}$ ​ + + (e) Best affine approximation:​ $\displaystyle{T(s) = -\frac{1}{2}(s - 1) + 1}$ + Equation of tangent line: $\displaystyle{y = -\frac{1}{2}(s - 1) + 1}$ + + + 3. (a) $f'(x = 4x$ + + %%(c)%% $\displaystyle{f'​(t) = -\frac{1}{2t^{\frac{3}{2}}}}$;​ $f$ is not differentiable at $t = 0$ + + (e) $y'(t) = 2t + 4$ + + + 4. Best affine approximation:​ $\displaystyle{T(t) = -\frac{1}{16}(x - 4) + \frac{1}{2}}$ + $\displaystyle{\frac{1}{\sqrt{3.98}} \approx T(3.98) = \frac{401}{800} = 0.50125}$ to five decimal places + + + 7. $T(t) = -9(t - 3) - 15$ + + + 11. $f'(0) = 0$ + + 13. $g$ is not differentiable at $x = 1$. + + + 14. (a) $\displaystyle{\frac{ds}{dt}} = 6t$ + + %%(c)%% $\displaystyle{\frac{dq}{ds} = 1 + \frac{2}{s^2}}$ ​ + + + 15. $\displaystyle{\frac{d}{dx}}(4x^2) = 8x$, $\displaystyle{\frac{d}{du}(3\sqrt{u-1}) = \frac{3}{2\sqrt{u-1}}}$ ​ + + + 16. (a) Average velocity over $[0, 1] = 0.2 \text{ meters/​second}$ ​ + + (b) $v = 10 - 9.8t \text{ meters/​second}$ + + %%(c)%% $v|_{t=1} = 0.2 \text{ meters/​second}$ + + (d) $v = 0$ when $t = 1.0204$ seconds. $v$ is positive before this time, and negative after. + + (e) $s|_{t=1.0204} = 105.10 \text{ meters}$, which must be the maximum height reached by the object. ​ + + (f) $\displaystyle{\frac{dv}{dt}} = -9.8 \text{ meters/​second}^2$ + + (g) The object is losing velocity at the same rate at all times. ​ + + + 17. $\displaystyle{\frac{dA}{dr}} = 2\pi r$ + + + 18. $\displaystyle{\frac{dV}{dr}} = 4\pi r^2$ + + + 19. $\displaystyle{\frac{dA}{dx}} = 2x$ + + ---- + + **Section 3.3** + + 1. (a) $f'(x) = 3x^2 + 6$ + + %%(c)%% $g'(t) = 3 - 12t$ + + (e) $f'(t) = 18t - 36$ + + + 2. (a) $f'(x) = 8x + 4$ + + %%(c)%% $\displaystyle{g'​(x) = \frac{11}{(2x + 5)^2}}$ ​ + + (e) $\displaystyle{f'​(t) = \frac{6t^6 + 104t^3 - 6t^2 - 48}{(2t^3 + 6)^2}}$ ​ + + (g) $h'(t) = -\displaystyle{\frac{3}{t^2}}$ + + (i) $\displaystyle{h'​(z) = 24z^2 + \frac{1}{2z^2}}$ ​ + + + 3. (a) $\displaystyle{\frac{ds}{dt}} = (t^2 - 6t +3)(32t^3 + 12t) + (8t^4 + 6t^2 - 7)(2t - 6)$ + + %%(c)%% $\displaystyle{\frac{dy}{dx}} = (x^2 - 2x + 3)(2x^2 + 13x - 6)(6x - 4) + (x^2 - 2x + 3)(4x + 13)(3x^2 - 4x + 1) + (2x - 2)(2x^2 + 13x - 6)(3x^2 - 4x + 1)$ + + + 4. (a) $k'(2) = 26$ + + %%(c)%% $k'(2) = 102$ + + + 5. (a) $\displaystyle{v(t) = -1 + \frac{1}{2}t^2}$ ​ + + %%(c)%% The object is at the origin when $t = -\sqrt{6}, 0, \sqrt{6}$. The velocities are: $v(-\sqrt{6}) = 2$, $v(0) = -1$, $v(\sqrt{6}) = 2$. + + (e) $-\sqrt{2} < t < \sqrt{2}$ + + (g) $a(t) = t$ + + (i) The object is moving toward the left and slowing down at time $t = 1.$ + + ---- + + **Section 3.4** + + 1. (a) $f'(x) =16(4x + 5)^3$ + + %%(c)%% ​ $\displaystyle{h'​(t) = -\frac{18}{(6t - 3)^3}}$ ​ + + (e) $g'(z) = 2(3z+4)^3(2z^2 + z)(4z + 1) + 9(3z + 4)^2(2z^2 + z)^2$ + + + + 2. (a) $\displaystyle{\frac{ds}{dt} = 16t^3(t^2 - 1) + 8t(t^2 - 1)^2}$ ​ + + + %%(c)%% $\displaystyle{\frac{dq}{dt} = \frac{9t^2 - 4}{2\sqrt{3t^3 - 4t}}}$ ​ + + + (e) $\displaystyle{\frac{dx}{dt} = \frac{40 - 32t}{(4t + 5)^2}}$ ​ + + + (g) $\displaystyle{\frac{dy}{dx} = \frac{3}{5(3x - 1)^{\frac{4}{5}}}}$ ​ + + + + 3. $\displaystyle{T(x) = -\frac{33}{125}(x - 2) + \frac{6}{25}}$ ​ + + + + 4. (a) $T(x) = hx + 1$ + + %%(c)%% $\displaystyle{\sqrt[3]{1.06} \approx \frac{1}{3}(0.06) + 1 = 1.02}$ + Rounded to four decimal places, $\sqrt[3]{1.06} = 1.0196$. + + + 5. (a) $\displaystyle{y = -\frac{1}{2}(x - 2) + 3}$ + + %%(c)%% $\displaystyle{y = -\frac{7}{8}(x - 2) + 1}$ + + (e) $y = 2 - x$ + + + 6. (a) $k'(0) = 9$ + + %%(c)%% $k'(0) = -6$ + + (e) $k'(0) = 8$ + + + 8. $900 \text{ cm}^3\text{/​sec}$ + + + 9. $400\pi \text{ cm}^2\text{/​sec}$ + + + 11. $\displaystyle{\frac{dK}{dt}\bigg|_{v=10} = 98m}$ + + + 13. $1.6 \text{ in}^2\text{/​sec}$ + + + 15. (a) $0$ units per month  ​ + + (b) $-100$ units per month + + + 17.$\displaystyle{\frac{5}{2\pi}} \text{ ft/sec}$ + + ---- + + **Section 3.5** + + 1. (a) $f'(x) = x^2\cos(x) + 2x\sin(x)$ + + %%(c)%% $g'(t) = -6t\sin(2t) + 3\cos(2t)$ + + (e) $f'(t) = -4\sin(3)t\sin(4t) + 3\cos(3t)\cos(4t)$ + + (f) $g'(z) = 12\sin^2(4z)\cos(4z)$ + + + 2. (a) $\displaystyle{\frac{dy}{dx} = \frac{2x\cos(2x) - \sin(2x)}{x^2}}$ ​ + + %%(c)%% $\dfrac{dx}{dt} = 8t\cos(4t^2 + 1)$ + + (e) $\dfrac{dz}{dt} = 2\sec(2t)\tan(2t)$ + + (g) $\dfrac{dy}{dx} = -2x^2\csc(2x)\cot(2x) + 2x\csc(2x)$ + + + 3. (a) $\dfrac{d}{dx}(\sin^2(2x)\cos^2(3x) = -6\sin^2(2x)\cos(3x)\sin(3x) + 4\cos^2(3x)\sin(2x)\cos(2x)$ ​ + + %%(c)%% $\dfrac{d}{dq}\sec^3(q^2) = 6q\sec^3(q^2)\tan(q^2)$ + + (e) $\displaystyle{\frac{d}{dz}\sqrt{1 + \sin^2(z)} = \frac{\sin(z)\cos(z)}{\sqrt{1 + \sin^2(z)}}}$ ​ + + + 5. $T(t) = 1$ + + + 7. (a) $S(x) = \dfrac{1}{2}x + 1$ + + (b) $T(x) = 4x$ + + %%(c)%% $U(x) = 2x + 1$ + + (e) $h = f \circ g$ and $U = S \circ T$ + + + 8. (a) $\displaystyle{\lim_{x \to 0}\frac{\sin(2x)}{x} = 2}$ + + %%(c)%% $\displaystyle{\lim_{x \to 0}\frac{\tan(x)}{x} = 1}$ + + (e) $\displaystyle{\lim_{x \to 0}\frac{\sin^2(x)}{x} = 0}$ + + (g) $\displaystyle{\lim_{t \to 0}\frac{\sin^2(3t)}{t^2} = 9}$ + + + 9. (a) $f$ is $O(h)$, but not $o(h)$. + + %%(c)%% $g$ is $O(h)$, but not $o(h)$. + + (e) $f$ is $o(h)$, and so also $O(h)$. + + + 11. (a) $D(0.00001) = 4.000010000027032$ to 15 decimal places + $D_1(0.00001) = 4.000000000026205$ to 15 decimal places + $D_2(0.00001) = 3.999999999848569$ to 15 decimal places + $f'(2) = 2$ + + %%(c)%% $D(0.00001) = 0.999999999983333$ to 15 decimal places + $D_1(0.00001) = 0.999999999983333$ to 15 decimal places + $D_2(0.00001) = 1.000000000000000$ to 15 decimal places + $f'(0) = 1$ + + ---- + + **Section 3.6** + + 1. (a) Three solutions: $-2.3407$, $-0.9018$, $2.4080$ + + + %%(c)%% Two solutions: $-0.8241$, $0.8241$ + + + (e) Two solutions $0.7135$, $2.0730$ + + + + 2. (d) If $\displaystyle{L = \lim_{n \to \infty}x_n}$,​ then + $\displaystyle{L = \lim_{n \to \infty}x_{n+1} = \lim_{n \to \infty}\frac{x_n + \dfrac{c}{x_n}}{2} = \frac{L + \dfrac{c}{L}}{2}}$,​ + from which it follows that $L^2 = c$. + + + + 3. $1.25992$ + + + + 4. $1.25850$ + + + + 6. The values oscillate between $1$ and $-1$; that is, for $n = 0, 1, 2, \ldots$, $x_{2n} = 1$ and $x_{2n + 1} = -1$. + + ---- + + **Section 3.7** + + 1. Let $g(x) = \cos(x) - x$. Then $g(0) = 1$ and $g(1) = \cos(1) - 1 < 0$ (since $\cos(1) < 1$). 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 + $f'​(c) = \frac{f(v) - f(u)}{v - u}.$ + 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 $(-\infty, 0)$. + + %%(c)%% $h$ is decreasing on both $(-\infty, 1)$ and $(1, \infty)$. There are no intervals on which $h$ is increasing. ​ + + (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 + $f'(c) = \frac{f(v) - f(u)}{v - u}.$ + 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 $k$. + + ​%%(c)%% $G(x) = -\cos(x) + k$ is an antiderivative of $g$ 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. $k = -1$ + + ---- + + **Section 3.8** + + 1. (a) $f$ has a maximum value of $12$ at $x = 4$ and a minimum value of $-4$ at $x = 0$. + + %%(c)%% $g$ has a maximum value of $\sqrt{2}$ at $t = -\dfrac{\pi}{4}$ and a minimum value of $-\sqrt{2}$ at $t = \dfrac{3\pi}{4}$. + + (e) $g$ has a maximum value of $0.5498$ at $x = -1.0678$ and at $x = 1.0678$, and a minimum value of $-1.6646$ at $x = -2$ and $x = 2$. + + (g) $f$ has a maximum value of $3.9453$ at $t = 2.2889$ and a minimum value of $0$ at $t = 0$ and $t = \pi$. + + + 3. The side parallel to the river should be $250$ yards long and the other two sides should be $125$ yards long. + + 5. $\$30.94$​ + + + 6. The total area is maximized when the entire length of the wire is used for the circle. The total area is minimized when a piece of length$ \dfrac{\pi}{4 + \pi} $is used for the circle and the remaining piece for the square. ​ + + + 7. (a)$ f $has local minimum of$ 5 $at$ x = 0 $.$ f $does not have a local maximum. ​ + + %%(c)%%$ g $has a local maximum of$ 4 $at$ t  = -2 $and a local minimum of$ 0 $at$ t = 0 $. + + (e)$ f $has a local maximum of$ 1 $at$ x = 0 $.$ f $does not have a local minimum. ​ + + (g)$ g $has a local maximum of$ \displaystyle{\frac{6}{25}\sqrt{\frac{3}{5}}}$at$ \displaystyle{x = -\sqrt{\frac{3}{5}}}$and a local minimum of$ \displaystyle{-\frac{6}{25}\sqrt{\frac{3}{5}}}$at$ \displaystyle{x = \sqrt{\frac{3}{5}}}$. ​ + + (i)$ g $has a local maximum of$ 0 $at$ t = 0 $and a local minimum of$ -\dfrac{256}{81} $at$ t = \dfrac{4}{3} $. + + + 8. (a)$ f''​(x) = 6 $+ + %%(c)%%$ \displaystyle{\frac{d^2s}{dt^2} = \frac{8}{(2t - 1)^3}}$​ + + (e)$ \displaystyle{\frac{d^2x}{dt^2} = -20\sin(2t)\cos(4t) - 16\cos(2t)\sin(4t)}$​ + + + 9.$ f $has a maximum value on$ (0, \infty) $of$ \dfrac{1}{4} $at$ x = 2 $.$ f $does not have a minimum value on$ (0, \infty) $. + + + 11.$ 10 \text{ feet} \times 10 \text{ feet} \times 10 \text{ feet} $+ + + 13. The base of the bin should be$ 11.45 \text{ feet} \times 11.45 \text{ feet} $and the height should be$ 7.63 \text{ feet} $. + + + 15.The radius of the can should be$ \displaystyle{\sqrt[3]{\frac{250}{\pi}}}$and the height should be$ \displaystyle{4\sqrt[3]{\frac{250}{\pi}}}$. ​ + + + 16. The radius of the can should be$ \sqrt[3]{62.5} $and the height should be$ \displaystyle{\frac{8}{\pi}\sqrt[3]{62.5}}$. ​ + + + 20. (b)$ L $is maximized when$ \displaystyle{\theta = \frac{11}{20}}$. The probabilty of$ AA $is$ 0.3025 $, the probability of$ Aa $is$ 0.4950 $, and the probabilty of$ aa $is$ 0.2025 $. + + + 22. (a)$ v(t) = 3\pi\cos(\pi t) $+ + (b)$ a(t) = -3\pi^2\sin(\pi t) $+ + %%(c)%% From time$ t = 0 $until time$ t = \dfrac{1}{2} $, the object moves to the right from$ x = 0 $to$ x = 3 $and is slowing down. From time$ t = \dfrac{1}{2} $until time$ t = 1 $, the object moves to the left from$ x = 3 $to$ x = 0 $and is speeding up. From time$ t = 1 $until time$ t = \dfrac{3}{2} $, the object moves to the left from$ x = 0 $to$ x = -3 $and is slowing down. From time$ t = \dfrac{3}{2} $until time$ t = 2 $, the object moves to the right from$ x = -3 $to$ x = 0 $and is speeding up. + + ---- + + **Section 3.9** + + 1. (a)$ f $is decreasing on$ \displaystyle{\left(-\infty,​ \frac{1}{2}\right)}$and increasing on$ \displaystyle{\left(\frac{1}{2},​ \infty\right)}$. The graph of$ f $is concave up on$ (-\infty, \infty) $.$ f $has a local minimum of$ -\dfrac{1}{4} $at$ x = \dfrac{1}{2} $. The graph of$ f $does not have any inflection points or asymptotes. ​ + + %%(c)%%$ g $is increasing on$ (-\infty, -2) $and$ (0, \infty) $, and decreasing on$ (-2, 0) $. The graph of$ g $is concave down on$ (-\infty, -1) $and concave up on$ (-1, \infty) $.$ g $has a local maximum of$ 4 $at$ x = -2 $and a local minimum of$ 0 $at$ x = 0 $.$ (-1, 2) $is an inflection point. The graph of$ g $does not have any asymptotes. ​ + + (e)$ f $is increasing on$ (-\infty, -1) $and$ (1, \infty) $, and decreasing on$ (-1, 1). $The graph of$ f $is concave down on$ (-\infty, 0) $and concave up on$ (0, \infty) $.$ f $has a local maximum of$2$at$ x = -1 $and a local minimum of$ -2 $at$ x = 1 $.$ (0, 0) $is an inflection point. The graph of$ f $does not have any asymptotes. ​ + + (g)$ h $is increasing on$ \displaystyle{\left(-\infty,​ -\sqrt{\frac{3}{5}}\right)}$and$ \displaystyle{\left(\sqrt{\frac{3}{5}},​ \infty\right)}$,​ and decreasing on$ \displaystyle{\left(-\sqrt{\frac{3}{5}},​ 0\right)}$and$ \displaystyle{\left(0,​ \sqrt{\frac{3}{5}}\right)}$. The graph of$ h $is concave down on$ \displaystyle{\left(-\infty,​ -\sqrt{\frac{3}{10}}\right)}$and$ \displaystyle{\left(0,​ \sqrt{\frac{3}{10}}\right)}$,​ and concave up on$ \displaystyle{\left(-\sqrt{\frac{3}{10}},​ 0\right)}$and$ \displaystyle{\left(\sqrt{\frac{3}{10}},​ \infty\right)}$.$ h $has a local maximum of$ \displaystyle{\frac{6}{25}\sqrt{\frac{3}{5}}}$at$ \displaystyle{x = -\sqrt{\frac{3}{5}}}$and a local minimum of$ \displaystyle{-\frac{6}{25}\sqrt{\frac{3}{5}}}$at$ \displaystyle{x = \sqrt{\frac{3}{5}}}$.$ \displaystyle{\left(-\sqrt{\frac{3}{10}},​ \frac{21}{100}\sqrt{\frac{3}{10}}\right)}$,​$ (0, 0) $, and$ \displaystyle{\left(\sqrt{\frac{3}{10}},​ -\frac{21}{100}\sqrt{\frac{3}{10}}\right)}$are inflection points. The graph of$ h $does not have any asymptotes. ​ + + (i)$ g $is decreasing on$ (-\infty, 1) $and$ (1, \infty) $. The graph of$ g $is concave down on$ (-\infty, 1) $and concave up on$ (1, \infty) $. The line$ z = 1 $is a vertical asymptote and the line$ y = 0 $(that is, the$ z $-axis) is a horizontal asymptote.$ g $does not have any local extreme values and the graph of$ g $does not have any inflection points. ​ + + (k)$f$is decreasing on$ (-\infty, 0) $and$ \displaystyle{\left(0,​ \frac{3}{2}\right)}$,​ and increasing on$ \displaystyle{\left(\frac{3}{2},​ 0\right)}$. The graph of$ f $is concave up on$ (-\infty, 0) $and$ (1, \infty) $, and concave down on$ (0, 1) $.$ f $has a local minimum of$ -\dfrac{27}{16} $at$ x = \dfrac{3}{2} $.$ (0,0) $and$ (1, -1) $are inflection points. The graph of$ f $does not have any asymptotes. ​ + + (m)$ h $is decreasing on$ (-\infty, -2) $,$ (-2, 2) $, and$ (2, \infty) $. The graph of$ h $is concave down on$ (-\infty, -2) $and$ (0, 2) $, and concave up on$ (-2, 0) $and$ (2, \infty) $.$ (0, 0) $is an inflection point. The lines$ t = -2 $and$ t = 2 $are vertical asymptotes and the line$ y = 0 $(that is, the$ t $-axis) is a horizontal asymptote.$ h $does not have any local extreme values. ​ + + (o)$ f $is decreasing on$ (-\infty, 0) $and increasing on$ (0, \infty) $. The graph of$ f $is concave down on$ \displaystyle{\left(-\infty,​ -\frac{1}{\sqrt{3}}\right)}$and$ \displaystyle{\left(\frac{1}{\sqrt{3}},​ \infty\right)}$,​ and concave up on$ \displaystyle{\left(-\frac{1}{\sqrt{3}},​ \frac{1}{\sqrt{3}}\right)}$.$ f $has a local minimum of$ 0 $at$ x = 0 $.$ \displaystyle{\left(-\frac{1}{\sqrt{3}},​ \frac{1}{4}\right)}$and$ \displaystyle{\left(\frac{1}{\sqrt{3}},​ \frac{1}{4}\right)}$are inflection points. The line$ y = 1 $is a horizontal asymptote. ​ + + (q)$ x $is decreasing on$ (-\infty, 1) $and$ (1, \infty) $. The graph of$ x $is concave down on$ (-\infty, 1) $and concave up on$ (1, \infty) $. The line$ t = 1 $is a vertical asymptote and the line$ y = 2 $is a horizontal asymptote.$ x $does not have any local extreme values. The graph of$ x \$ does not have any inflection points. ​ + + + + 3. %%(c)%% There is only one function which satisfies these conditions. ​ + + + 4. %%(c)%% There is only one function which satisfies these conditions. ​ + + + 5. %%(c)%% There is only one function which satisfies these conditions. + + ----