Science:Math Exam Resources/Courses/MATH104/December 2016/Question 05 (a)

MATH104 December 2016

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Question 05 (a)
Find the absolute minimum $y=x{\sqrt[{3}]{x-3}}=x(x-3)^{1/3}$ on the interval $\left[0,4\right]$

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Hint
Find the critical numbers, and then compare these values with 2 end points.

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First, we need the critical numbers. To this end, we first find the derivative of $f$ ; ${\begin{aligned}f'(x)&=(x-3)^{1/3}+x\cdot {\frac {1}{3}}(x-3)^{-2/3}\\&=(x-3)^{1/3}+{\frac {x}{3(x-3)^{2/3}}}\\&={\frac {3(x-3)+x}{3(x-3)^{2/3}}}\\&={\frac {4x-9}{3(x-3)^{2/3}}}.\end{aligned}}$ Since $f'\left({\frac {9}{4}}\right)=0$ and $f'(x)$ is not defined at $x=3$ , the critical numbers are $x={\frac {9}{4}}$ and $x=3$ . We can now test these values and the 2 end points: $f(0)=f(3)=0$ $f(4)=4\cdot 1=4$ $f\left({\frac {9}{4}}\right)={\frac {9}{4}}\cdot {\sqrt[{3}]{-3/4}}<0$ Therefore the absolute minimum of $f$ on the interval $[0,4]$ is $\color {blue}f\left({\frac {9}{4}}\right)={\frac {9}{4}}\cdot {\sqrt[{3}]{-3/4}}$ .