## Calculus: Early Transcendentals 8th Edition

(a) $a = 0$ $b =-1$ (b) The tangent line with the smallest slope is $~~y = -x$
(a) $f(x) = x^3+ax^2+bx$ $f'(x) = 3x^2+2ax+b$ It is given that $f$ has a local minimum at $x = \frac{1}{\sqrt{3}}$: $f'(\frac{1}{\sqrt{3}}) = 0$ $f'(\frac{1}{\sqrt{3}}) = 3(\frac{1}{\sqrt{3}})^2+2a(\frac{1}{\sqrt{3}})+b = 0$ $1+\frac{2a}{\sqrt{3}}+b = 0$ $b = -1-\frac{2a}{\sqrt{3}}$ It is given that $f(\frac{1}{\sqrt{3}}) = -\frac{2}{9}\sqrt{3}$: $f(\frac{1}{\sqrt{3}}) = (\frac{1}{\sqrt{3}})^3+a(\frac{1}{\sqrt{3}})^2+b(\frac{1}{\sqrt{3}}) = -\frac{2}{9}\sqrt{3}$ $\frac{1}{3\sqrt{3}}+\frac{a}{3}+\frac{b}{\sqrt{3}} = -\frac{2}{9}\sqrt{3}$ $\frac{1}{3\sqrt{3}}+\frac{a}{3}+\frac{(-1-\frac{2a}{\sqrt{3}})}{\sqrt{3}} = -\frac{2}{9}\sqrt{3}$ $\frac{1}{3\sqrt{3}}+\frac{a}{3}-\frac{1}{\sqrt{3}}-\frac{2a}{3} = -\frac{2}{9}\sqrt{3}$ $\frac{a}{3}-\frac{2a}{3} = -\frac{2}{9}\sqrt{3}-\frac{1}{3\sqrt{3}}+\frac{1}{\sqrt{3}}$ $-\frac{a}{3} = -\frac{2}{9}\sqrt{3}-\frac{1}{3\sqrt{3}}+\frac{1}{\sqrt{3}}$ $a = \frac{2}{3}\sqrt{3}+\frac{1}{\sqrt{3}}-\sqrt{3}$ $a = \frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\sqrt{3}$ $a = \frac{3}{\sqrt{3}}-\sqrt{3}$ $a = \sqrt{3}-\sqrt{3}$ $a =0$ We can find the value of $b$: $b = -1-\frac{2a}{\sqrt{3}}$ $b = -1-\frac{2(0)}{\sqrt{3}}$ $b = -1$ Therefore: $f(x) = x^3-x$ $a = 0$ $b =-1$ (b) We can find the point where $f''(x) = 0$: $f'(x) = 3x^2+2ax+b = 3x^2-1$ $f''(x) = 6x = 0$ $x = 0$ $f'(x)$ has a minimum at $x = 0$ $f'(0) = 3(0)^2-1 = -1$ The slope of the tangent line at the point $x=0$ is $m = -1$ $f$ has the smallest slope when $x = 0$ $f(0) = (0)^3-(0) = 0$ We can find the equation of the tangent line at the point $(0,0)$: $y-0 = -1(x-0)$ $y=-x$ The tangent line with the smallest slope is $y = -x$