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

This function is only defined for x \geq 0. Same for part (b). In the hint, please add that points where the derivative DNE are also considered critical points. As a consequence, 0 is also a critical point here, since f'(x) \to \infty as x \to 0.

Pardon my ignorance, but why is the function not defined for x < 0? At x = -1, is its value not 77? $(-1)^{\frac {2}{7}}={\sqrt[{7}]{(-1)^{2}}}={\sqrt[{7}]{1}}=1$ and $(-1)^{\frac {9}{7}}={\sqrt[{7}]{(-1)^{9}}}={\sqrt[{7}]{-1}}=-1$. $63(1)-14(-1)=77$. Please help me to understand if I am overlooking something.

You are absolutely right! This works because of the odd denominator in the fractional power. I was confused because, in general, the n-th root of a negative number has n solutions, if you allow complex numbers. This is also the reason why google & wolframalpha etc tell you that (-1)^(2/7) is a complex number. None of that fanciness is necessary here, and your solution is very clear. Good job.