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I am analyzing this function: \begin{align} f(x) = \sqrt[3]{ x } - \sqrt[3]{x+1} \end{align}

I proved it has a horizontal asymptote at y=0. However, on the graph, it has a minimum for x=-0.5 but I can't find that. First derivative has no zeroes, so the function has no stacionary points. Second derivative also has no zeroes. How do I find that minimum?

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    i think the minimum is at the point $x=0$ – 2017-01-06
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    It's not. For x=0, y=-1 and for x=-0.5, y=-1.587. – 2017-01-06
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    The first derivative does have zero at x=-1/2 – 2017-01-06
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    If others did what I did and typed the expression into wolfram alpha, it will assume you want the principle root and plot real and complex parts, which is what makes it look like the minimum is at zero. – 2017-01-06

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You must've taken the first derivative wrong:

$$f'(x)=\frac13(x^{-2/3}-(x+1)^{-2/3})$$

Setting $f'(x)=0$,

$$0=x^{-2/3}-(x+1)^{-2/3}$$

$$x^{-2/3}=(x+1)^{-2/3}$$

$$\pm x=x+1$$

$$x=-1/2$$

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    Ahh, I forgot the + and - at the third step of equation. – 2017-01-06
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    @Nebeski Yes, that $\ ^2$ sure gets to you :D – 2017-01-06
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Don't forget that critical points also occur where the derivative is undefined, as long as the point in question is in the domain of the original function.

Since $f(x) = x^{1/3} - (x+1)^{1/3}$, then $f'(x) = \dfrac13 x^{-2/3} - \dfrac13 (x+1)^{-2/3}$.

Note that $f'(x)$ is undefined at $x=0$ and $x=-1$, and both of these are in the domain of $f(x)$, so there may be extrema at those locations.

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    Why is it undefined there? Cube root of zero is zero. – 2017-01-06
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    @Nebeski, note the negative exponents. For example, $\dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}$. Let $x = 0$ and what happens? – 2017-01-06
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    @Nebeski You get division by $0$. Note the shape of the $y=\sqrt[3]{x}$ graph around $x=0$. – 2017-01-06
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    @tilper No worries, I mislooked. – 2017-01-06
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    (although there are *not* extrema at those values) – 2017-01-06
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    Yeah, it should be obvious that $f(x)$ is continuous and never diverges... – 2017-01-06
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    I was looking at the function, not the derivative. I didn't know I should look where the first derivative is defined. I would just find x so derivative is equal to zero, and then check those points. – 2017-01-06
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    @ElliotG, true, but the other answer already explained how to find the CP, and OP appeared not to know that CP can also occur where $f'$ is undefined so I saw this as a teachable moment. – 2017-01-06
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    @Nebeski, always check where the first derivative is zero and where it's undefined (but make sure you always stay in the domain of the original function). For example, $g(x) = x^{2/3}$ has a minimum at $x=0$ and $g'(x)$ is not defined at $x=0$. – 2017-01-06
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    always a good reminder :) – 2017-01-06