Science:Math Exam Resources/Courses/MATH100/December 2011/Question 05 (a)

MATH100 December 2011

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Question 05 (a)
Full-Solution Problems. In questions 2-8, justify your answers and show all your work. Simplification of answers is not required unless explicitly stated. Let $\displaystyle y=f(x)=6x^{1/5}+x^{6/5}.$ Find the critical numbers of y = ƒ(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
Critical points (numbers) are roots of the derivative.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Critical points (numbers) are roots of the derivative. Let's compute this derivative: $f'(x)={\frac {6}{5}}x^{-4/5}+{\frac {6}{5}}x^{1/5}$ And now find the roots: ${\frac {6}{5}}x^{-4/5}+{\frac {6}{5}}x^{1/5}=0$ We can divide by 6/5 $\displaystyle x^{-4/5}=-x^{1/5}$ Now, clearly x = 0 is a solution. To find possibly others, we now assume that x isn't zero and now multiply by x4/5 $\displaystyle 1=-x$ And obtain a second solution. So we can conclude that there are two critical points: x = 0 and x = -1.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Critical points (numbers) are roots of the derivative. Let's compute this derivative: $f'(x)={\frac {6}{5}}x^{-4/5}+{\frac {6}{5}}x^{1/5}$ And now we can do a little algebra on the function and rewrite as ${\begin{aligned}f'(x)&={\frac {6}{5}}x^{-4/5}+{\frac {6}{5}}x^{1/5}\\&={\frac {6}{5}}x^{1/5}(x^{-1}+1)\end{aligned}}$ And so the critical points are solutions of either $\displaystyle x^{1/5}=0\quad \iff \quad x=0$ or $\displaystyle x^{-1}+1=0\quad \iff \quad x=-1$ And so we have two critical points: x = 0 and x = -1.

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