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de2de:chapter-3 [2012/06/25 11:25] (current)
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 +====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.
 +
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