Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone show me how to simplify get the critical point of this function.


I did the product rule and got


but I am having touble simplifying such this how would I simplify it can anyone show me how this would be done.

share|cite|improve this question
up vote 4 down vote accepted

You need only find the points at which $f'(x) = 0$, and where $f'(x)$ is undefined.

To simplify, find a common denominator and add terms, then set equal to zero: The common denominator is $3\sqrt[\large 3]{(2 + x)^2}$. Do you recall how to bring all terms over this common denominator?

We have $$2x\sqrt[\large3]{2+x}+\frac{x^2}{3\sqrt[\large3]{(2+x)^2}}$$

And want

$$ \begin{align} f'(x) & = \dfrac{2x \sqrt[\large 3]{2+x}\times 3\sqrt[\large 3]{(2+x)^2} + x^2}{3\sqrt[\large 3]{(2+x)^2}} \\ \\ & = \frac{6x \sqrt[\large 3]{(2+x)(2+x)^2} + x^2}{3 \sqrt[\large 3]{(2+x)^2}} \\ \\ & = \frac{6x \sqrt[\large 3]{(2+x)^3} + x^2}{3 \sqrt[\large 3]{(2+x)^2}} \\ \\ & = \dfrac{6x(2 + x) + x^2}{3\sqrt[\large 3]{(2+x)^2}} \\ \\ & = \dfrac{12x + 6x^2 + x^2}{3 \sqrt[\large 3]{(2+x)^2}} \\ \\ & = \dfrac{x(12 + 7x)}{3\sqrt[\large 3]{(2 + x)^2}} = 0 \\ \\ \end{align} $$

$f'(x)$ is undefined when $x = -2$.

$f'(x) = 0$ when the numerator equals zero: one point at which this occurs is when $x = 0$.

$f'(x) = 0$ when $(12 + 7x) =0 \iff 7x = -12 \iff x = -\large\frac{12}{7}$

Three critical points in all. The blue line below is your function of interest. Note the sharp corner at $x = -2$. It happens to be a local minimum. Also note the local maximum at $x = -\large\frac{12}{7}$, and the minimum at $x = 0$. Graphs are really helpful to confirm the work you're doing, and better understand the behavior of the function.

enter image description here

ASIDE: Personally I think using fractional exponents to express roots makes this sort of problem a bit clearer, in terms of algebraic manipulation, particularly when we're talking about roots other than the square root, and especially when they appear in fractions.

share|cite|improve this answer
Nice Ammy. :.+) – Babak S. Mar 5 '13 at 19:49
thanks for the help. – Fernando Martinez Mar 7 '13 at 15:32

On simplification $$f'(x)=\frac{6x(2+x)^{3/3}+x^2}{3(2+x)^{2/3}}=\frac{12x+7x^2}{3(2+x)^{2/3}}$$

Now, critical poits are where $f'(x)=0$ which gives $x=0,x=-12/7$

and those points where $f'(x)$ is not defined like $x=-2$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.