Calculus: Early Transcendentals 8th Edition

(a) Using the graph, we can estimate that the maximum value is approximately $f(1) = 1.4$ Using calculus, we can find that the maximum value is $f(1) = \sqrt{2}$ (b) The graph appears to increase most rapidly, that is it has the highest positive slope, approximately at the point $x = -0.5$ Using calculus, the slope of the graph is a maximum when $x = \frac{3-\sqrt{17}}{4} \approx -0.28$
(a) $f(x) = \frac{x+1}{\sqrt{x^2+1}}$ Using the graph, we can make the following estimates: The maximum value is approximately $f(1) = 1.4$ There is no minimum value but $\lim\limits_{x \to -\infty}f(x) = -1$ We can find the exact values: $f'(x) = \frac{\sqrt{x^2+1}-(x+1)(\frac{x}{\sqrt{x^2+1}})}{x^2+1}$ $f'(x) = \frac{\frac{x^2+1}{\sqrt{x^2+1}}-(\frac{x^2+x}{\sqrt{x^2+1}})}{x^2+1}$ $f'(x) = \frac{1-x}{(x^2+1)^{3/2}} = 0$ $1-x = 0$ $x = 1$ $f(1) = \frac{1+1}{\sqrt{1^2+1}} = \frac{2}{\sqrt{2}} = \sqrt{2}$ The maximum value is $f(1) = \sqrt{2}$ There is no minimum value but we can find $\lim\limits_{x \to -\infty}f(x)$ $\lim\limits_{x \to -\infty} \frac{x+1}{\sqrt{x^2+1}} = \lim\limits_{x \to -\infty} \frac{x/x+1/x}{-\sqrt{x^2/x^2+1/x^2}} = -1$ (b) The graph appears to increase most rapidly, that is it has the highest positive slope, approximately at the point $x = -0.5$ We can find $x$ when $f''(x) = 0$: $f''(x) = \frac{(-1)(x^2+1)^{3/2}-(1-x)(\frac{3}{2})(\sqrt{x^2+1})(2x)}{(x^2+1)^3}$ $f''(x) = \frac{-(x^2+1)^{3/2}-(3x-3x^2)(\sqrt{x^2+1})}{(x^2+1)^3} = 0$ $-(x^2+1)^{3/2}-(3x-3x^2)\sqrt{x^2+1} = 0$ $(x^2+1)^{3/2} = (3x^2-3x)\sqrt{x^2+1}$ $x^2+1 = 3x^2-3x$ $2x^2-3x-1 = 0$ We can use the quadratic formula: $x = \frac{3 \pm \sqrt{(-3)^2-(4)(2)(-1)}}{2(2)}$ $x = \frac{3 \pm \sqrt{9+8}}{4}$ $x = \frac{3 \pm \sqrt{17}}{4}$ Using the graph, we can clearly see that the negative branch is the correct solution. The slope of the graph is a maximum when $x = \frac{3-\sqrt{17}}{4} \approx -0.28$