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I need to study this function:

$$f(x) = \frac{\sqrt[3]{x-1}}{(x+2)^2}$$

and I need to show Max and Min point.

The first thing is define the Domain, so:

$$\left\{\begin{matrix} \sqrt[3]{x-1} > 0\\ (x+2)^2 \neq 0 \end{matrix}\right.$$

$$\left\{\begin{matrix} x > 1\\ x \neq -2 \end{matrix}\right.$$

So my domain is:

$$(1, +\infty )$$

Now I check the intersection with x and y:

$$\frac{\sqrt[3]{x-1}}{(x+2)^2} = 0$$

and I get $$ N: x = 1 $$ $$ D: x = -2 $$

And I have no intersection with y. I have checked the limit in $1$ and $+\infty$:

$$ \lim_{x \rightarrow +\infty} \frac{\sqrt[3]{x-1}}{(x+2)^2} = 0$$ $$ \lim_{x \rightarrow 1^+} \frac{\sqrt[3]{x-1}}{(x+2)^2} = 0$$ $$ \lim_{x \rightarrow 1^-} \frac{\sqrt[3]{x-1}}{(x+2)^2} = 0$$

I have calculated the first derivative as suggested here, the result is:

$$ f'(x) = \frac{8-5x}{3\sqrt[3]{(x-1)^2}(x+2)}$$

Now how should I precede to study Max and Min? Are my steps correct?

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Christian Giupponi
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    Note: The cube root has domain $(-\infty,\infty)$ – Teoc Jun 25 '15 at 15:25
  • what happens if $x=1$? – user190080 Jun 25 '15 at 15:28
  • @VladimirLenin that is not true in general and pretty much depends on the definition, usually one asks x to be non-negative as far as i know – user190080 Jun 25 '15 at 15:30
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    That is only for square root, because $\sqrt {-x}=i\sqrt x$. But as you can see, $(-2)^3=-8$, so $\sqrt[3]{-8}=-2$. Also, $\sqrt[3] 0=0$. See the plot here: http://www.wolframalpha.com/input/?i=cbrt(x) – Teoc Jun 25 '15 at 15:34
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    @VladimirLenin I know that $(-2)^3=-8$, but that's not the point, you can also write that the solution to the equation $x^3=-8$ is $x=-\sqrt[3]{8}$, but you run into problems if you allow also negatives...what I want to say is, that Christian Giupponi should check his definition because there are two ways of defining it – user190080 Jun 25 '15 at 15:38
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    @user190080 there's no two ways of defining a cube root of a real number. Even if it had been a square root of a real number, definition is clear: √ is just positive. Positive-negative things just happen only with even roots. In cube root, it is clear that $\sqrt[3]{-8}$ is just a plain number $-2$. – Jeong Jinmyeong Jun 25 '15 at 15:48
  • so the root should be $\neq0$ instead of > ? – Christian Giupponi Jun 26 '15 at 13:27

1 Answers1

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First, you must review domain of your function. Why do you need $ \sqrt[3]{x-1} > 0$? Cube root is different from square root, it can have negative value without concerning about complex numbers, as @Vladimir Lenin pointed that out in comments.

Then you can now understand you need to check limits of the given function when $x \to -2$, not only $x \to 1$. Of course $x \to - \infty$ too. Then you'll understand what the function looks like.

Also you need to review your first derivative. The derivative you wrote down is not right.

Jeong Jinmyeong
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