Science:Math Exam Resources/Courses/MATH100/December 2013/Question 08 (a)

MATH100 December 2013

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Question 08 (a)
Full-Solution Problem. Justify your answer and show your work. Full simplification of numerical answers is required unless explicitly stated otherwise. Let $\displaystyle y=f(x)=63x^{2/7}-14x^{9/7}$. Find the critical numbers of $\displaystyle y=f(x)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
$x$ is said to be a critical number/point of $f$ if the slope of the tangent line to $f$ at $x$ is 0, or if the slope does not exist.

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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. By the definition of a critical number, we must find the $x$ for which $\displaystyle f'(x)=0$ or for which $\displaystyle f'(x)$ does not exist: $$f(x)=63x^{\frac {2}{7}}-14x^{\frac {9}{7}}$$ $$f'(x)={\frac {2}{7}}\cdot 63x^{{\frac {2}{7}}-1}-{\frac {9}{7}}\cdot 14x^{{\frac {9}{7}}-1}=18x^{-{\frac {5}{7}}}-18x^{\frac {2}{7}}=18x^{-{\frac {5}{7}}}(1-x)$$ Setting $\displaystyle f'(x)=0$ and applying the zero product property: $$18x^{-{\frac {5}{7}}}(1-x)=0\implies 18x^{-{\frac {5}{7}}}=0,\quad 1-x=0\implies x=1.$$ Note that $18x^{-{\frac {5}{7}}}$ is never equal to zero on the interval $(-\infty ,\infty )$. However, $18x^{-{\frac {5}{7}}}$ is undefined at $x=0$, so $\displaystyle f'(x)$ does not exist at $x=0$. Therefore the critical numbers of $f$ are $x={\color {blue}0,1}$.

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