How to solve $4(3x^3 -2000)/x^2 = 0$?

Question

The equation is $4(3x^3 -2000)/x^2 = 0$, and the function on the left side of the equation is the derivative of $f(x) = 6x^2 + 8000/x$. I'm looking for the local minimum. A computer gives the answer $(10/3) \sqrt[3]{18}$. I'm not supposed to use a calculator or a CAS, so I'm puzzled as to which strategy I could find here. I do know the shape of the graph of $f(x)$ and I can see how the computer answer makes sense. So the question is --- is there an easy strategy anywhere that I can't seem to notice?

Answer 1

Okay, so we take the function $f(x)=6x^2+\frac{8000}{x}$. Differentiating this yields:

$$f’(x)=12x-\frac{8000}{x^2}$$

Setting this to $0$ and multiplying by $x^2$, this becomes:

$$0=12x^3-8000,\text{ }x\neq 0$$

Which is a relatively simple algebra problem. Then use the second derivative test to determine what kind of extremum that point is.

You took your derivative wrong, hence why you got a different function for $f’(x)$. The power rule states that:

$$\frac{d}{dx}(x^a)=ax^{a-1}$$

For all $a\in\mathbb{R}$.