question 6
this is what we know so far:
function: ( f ( x ) = sqrt 3 { x ^ { 2 } - 6 x + 5 } )
domain: ( ( - infty , infty ) )
intercepts: ( ( 0 , sqrt 3 { 5 } ) ) and ( ( 1,0 ) ) and ( ( 5,0 ) )
asymptotes: none
derivative: ( f ^ { prime } ( x ) = \frac { 3 ( x - 1 ) ^ { \frac { 2 } { 3 } } ( x - 5 ) ^ { \frac { 2 } { 3 } } } { 2 x - 6 } )
critical points: ( x = 1 , x = 3 ), and ( x = 5 )
now complete the table below to find the monotonicity intervals of ( f ( x ) ) (on some devices/browsers
you may need to use the keyboard and the tab - key to navigate to certain answer fields, they may not be
accessible with a mouse or trackpad):

Question

question 6
this is what we know so far:
function: ( f ( x ) = sqrt  3  { x ^ { 2 } - 6 x + 5 } )
domain: ( ( - infty , infty ) )
intercepts: ( ( 0 , sqrt  3  { 5 } ) ) and ( ( 1,0 ) ) and ( ( 5,0 ) )
asymptotes: none
derivative: ( f ^ { prime } ( x ) = \frac { 3 ( x - 1 ) ^ { \frac { 2 } { 3 } } ( x - 5 ) ^ { \frac { 2 } { 3 } } } { 2 x - 6 } )
critical points: ( x = 1 , x = 3 ), and ( x = 5 )
now complete the table below to find the monotonicity intervals of ( f ( x ) ) (on some devices/browsers
you may need to use the keyboard and the tab - key to navigate to certain answer fields, they may not be
accessible with a mouse or trackpad):

Accepted Answer

# Explanation: ## Step1: Choose test points in each interval For the interval $(-\infty,1)$, let $x = 0$. $$ \begin{align*} f^{\prime}(0)&=\frac{3(0 - 1)^{\frac{2}{3}}(0 - 5)^{\frac{2}{3}}}{2(0)-6}\ &=\frac{3(- 1)^{\frac{2}{3}}(-5)^{\frac{2}{3}}}{-6}\ &=\frac{3(1)(\sqrt[3]{25})}{-6}\ &=-\frac{\sqrt[3]{25}}{2}<0 \end{align*} $$ Since $f^{\prime}(x)<0$ on $(-\infty,1)$, the function $f(x)$ is decreasing on $(-\infty,1)$. ## Step2: For the interval $(1,3)$, let $x = 2$ $$ \begin{align*} f^{\prime}(2)&=\frac{3(2 - 1)^{\frac{2}{3}}(2 - 5)^{\frac{2}{3}}}{2(2)-6}\ &=\frac{3(1)^{\frac{2}{3}}(-3)^{\frac{2}{3}}}{4 - 6}\ &=\frac{3(\sqrt[3]{9})}{-2}\ &=-\frac{3\sqrt[3]{9}}{2}<0 \end{align*} $$ Since $f^{\prime}(x)<0$ on $(1,3)$, the function $f(x)$ is decreasing on $(1,3)$. ## Step3: For the interval $(3,5)$, let $x = 4$ $$ \begin{align*} f^{\prime}(4)&=\frac{3(4 - 1)^{\frac{2}{3}}(4 - 5)^{\frac{2}{3}}}{2(4)-6}\ &=\frac{3(3)^{\frac{2}{3}}(-1)^{\frac{2}{3}}}{8 - 6}\ &=\frac{3(\sqrt[3]{9})(1)}{2}\ &=\frac{3\sqrt[3]{9}}{2}>0 \end{align*} $$ Since $f^{\prime}(x)>0$ on $(3,5)$, the function $f(x)$ is increasing on $(3,5)$. ## Step4: For the interval $(5,\infty)$, let $x = 6$ $$ \begin{align*} f^{\prime}(6)&=\frac{3(6 - 1)^{\frac{2}{3}}(6 - 5)^{\frac{2}{3}}}{2(6)-6}\ &=\frac{3(5)^{\frac{2}{3}}(1)^{\frac{2}{3}}}{12 - 6}\ &=\frac{3(\sqrt[3]{25})}{6}\ &=\frac{\sqrt[3]{25}}{2}>0 \end{align*} $$ Since $f^{\prime}(x)>0$ on $(5,\infty)$, the function $f(x)$ is increasing on $(5,\infty)$. # Answer: | Intervals | Sign of $f^{\prime}$ | Monotonicity | |---|---|---| | $(-\infty,1)$ | Negative ($<0$) | Decreasing | | $(1,3)$ | Negative ($<0$) | Decreasing | | $(3,5)$ | Positive ($>0$) | Increasing | | $(5,\infty)$ | Positive ($>0$) | Increasing |

Intervals	Sign of \(f^{\prime}\)	Monotonicity
\((1,3)\)	Negative (\(<0\))	Decreasing
\((3,5)\)	Positive (\(>0\))	Increasing
\((5,\infty)\)	Positive (\(>0\))	Increasing

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QUESTION IMAGE

Question

Explanation:

Step1: Choose test points in each interval

Step2: For the interval \((1,3)\), let \(x = 2\)

Step3: For the interval \((3,5)\), let \(x = 4\)

Step4: For the interval \((5,\infty)\), let \(x = 6\)

Answer: