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

$f(x)=x^2\sqrt[3]{2+x}$

I did the product rule and got

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

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

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2 Answers

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.

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Nice Ammy. :.+) –  B. S. Mar 5 '13 at 19:49
    
thanks for the help. –  Fernando Martinez Mar 7 '13 at 15:32
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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 $x=-2$

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