Question

question 5
this is what we know so far:
function: $f(x)=\sqrt3{x^{3}-6x + 5}$
domain: $(-\infty,\infty)$
intercepts: $(0,\sqrt3{5})$ and $(1,0)$ and $(5,0)$
asymptotes: none
derivative: $f(x)=\frac{2x - 6}{3(x - 1)^{\frac{2}{3}}(x - 5)^{\frac{2}{3}}}$
now enter the critical points below (in order from small to large):
$x=\square$
and $x=\square$
and $x=\square$

Explanation:

Step1: Find where the derivative is zero

Set \(f^{\prime}(x) = 0\). Since \(f^{\prime}(x)=\frac{2x - 6}{3(x - 1)^{\frac{2}{3}}(x - 5)^{\frac{1}{2}}}\), set the numerator equal to zero (\(2x-6 = 0\)).
\(2x-6=0\Rightarrow2x = 6\Rightarrow x = 3\)

Step2: Find where the derivative is undefined

The derivative \(f^{\prime}(x)\) is undefined when the denominator is zero.
Set \(3(x - 1)^{\frac{2}{3}}(x - 5)^{\frac{1}{2}}=0\).
For \((x - 1)^{\frac{2}{3}}=0\), \(x = 1\) (because \(y = u^{\frac{2}{3}}=\sqrt[3]{u^{2}}\), and when \(u=x - 1 = 0\), \(x = 1\)).
For \((x - 5)^{\frac{1}{2}}=0\), \(x = 5\) (because \(y=\sqrt{u}\), and when \(u=x - 5=0\), \(x = 5\))

Answer:

\(x = 1\) and \(x = 3\) and \(x = 5\)