Answer
(a) $a = 0$
$b =-1$
(b) The tangent line with the smallest slope is $~~y = -x$
Work Step by Step
(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 $y = -x$