How to solve this question

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mimie

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Given that [MATH]\text{f(}x\text{)}={{x}^{\frac{1}{3}}}[/MATH]. Find the value(s) of x when [MATH]\text{f '(}x\text{)}=0[/MATH] and state the interval(s) for which [MATH]\text{f '(}x\text{)}>0[/MATH] and [MATH]\text{f '(}x\text{)}<0[/MATH].

This is my calculation,
[MATH] \begin{align} \text{f (}x\text{)}&={{x}^{\frac{1}{3}}} \\ \text{f '(}x\text{)}&=\frac{1}{3{{x}^{\frac{2}{3}}}} \\ \text{when f '(}x\text{)}&=0 \\ \frac{1}{3{{x}^{\frac{2}{3}}}}&=0 \\ 1&=0 \end{align} [/MATH]Been stuck on this part, can anyone guide me? Thanks.
 
You have shown that no value of [MATH]x[/MATH] makes [MATH]f'(x)=0[/MATH], which is correct.
 
Look at the graph, and think about what it implies for your question. You'll find that it agrees with the next step of your work!
 
A fraction = 0 when the numerator is 0 AND the denominator is not 0. As you noted, your fraction equals 0 precisely when 1=0, which is never.

Please post back with the rest of your work. Thanks!
 
[MATH] \begin{align} \text{f (}x\text{)}&={{x}^{\frac{1}{3}}} \\ \text{f '(}x\text{)}&=\frac{1}{3{{x}^{\frac{2}{3}}}} \\ \text{when f '(}x\text{)}&=0 \\ \frac{1}{3{{x}^{\frac{2}{3}}}}&=0 \\ 1&=0 \end{align} [/MATH][MATH]\therefore since\ \frac{1}{3{{x}^{\frac{2}{3}}}}\ne 0 \text{ for all values of } x, \text{ so }x \text{ does not exist when f '(}x\text{)}=0. [/MATH]
I make a conclusion like this, but not sure if I write it correctly.

For the interval part, I want use first derivative test in the table form, but the solution of x when f '(x)=0 does not exist,
then what value should I use in first derivative test? Can someone guide me? Thanks.
 
mimie said:
[MATH] \begin{align} \text{f (}x\text{)}&={{x}^{\frac{1}{3}}} \\ \text{f '(}x\text{)}&=\frac{1}{3{{x}^{\frac{2}{3}}}} \\ \text{when f '(}x\text{)}&=0 \\ \frac{1}{3{{x}^{\frac{2}{3}}}}&=0 \\ 1&=0 \end{align} [/MATH][MATH]\therefore since\ \frac{1}{3{{x}^{\frac{2}{3}}}}\ne 0 \text{ for all values of } x, \text{ so }x \text{ does not exist when f '(}x\text{)}=0. [/MATH]
I make a conclusion like this, but not sure if I write it correctly.

For the interval part, I want use first derivative test in the table form, but the solution of x when f '(x)=0 does not exist,
then what value should I use in first derivative test? Can someone guide me? Thanks.
Click to expand...
My conclusion line would have been:

Since \(\displaystyle \lim_{x \to 0} \left[ \frac{1}{3*x^{\frac{2}{3}}}\right]\) does not exist, we conclude that f'(0) does not exist at x = 0.

For the interval part, I would first plot y = \(\displaystyle \left[ \frac{1}{3*x^{\frac{2}{3}}}\right] \) then decide on table format.
 
Since \(\displaystyle x^2\) is never negative, \(\displaystyle \frac{1}{3x^{2/3}}\) is never negative. Since, as you have already seen, it is never 0, \(\displaystyle f'(x)= \frac{1}{3x^{2/3}}\) is always positive.
 
mimie said:
For the interval part, I want use first derivative test in the table form, but the solution of x when f '(x)=0 does not exist,
then what value should I use in first derivative test? Can someone guide me? Thanks.
Click to expand...
What you found is that f'(x) is never zero; what you didn't mention is that f'(x) is undefined at x=0. Putting those together, the only place where the derivative might change sign is at x=0. So one approach you could take is to try a positive value and a negative value for x; you'll find it is positive at both, so f'(x)>0 on two intervals, x<0 and x>0. (You can't just say all real numbers, because it is not defined everywhere.)

Of course, other answers have shown other ways to get the answer.
 
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