# Local minimum and maximum points of $f(x)= \frac{x^2}{x-5}$ [closed]

Trying to find the local minimum and maximum points of the function

$$f(x)=\frac{x^2}{x-5}.$$

I have tried desmos calculator and notes.

• downvoted because it's not clear if you want x-5 or x^2/(x-5) – Alex Li Aug 28 '17 at 1:27
• Are you allowed to use methods from calculus? – platty Aug 28 '17 at 1:28
• Calculus is allowed – beccap1990 Aug 28 '17 at 1:31
• Hint (without calculus): $f(x) = 5\left(\frac{5}{x-5}+\frac{x-5}{5}\right)+10\,$. Let $u=\frac{x-5}{5}$ and consider the extrema of $u + \frac{1}{u}\,$. – dxiv Aug 28 '17 at 1:38
• – lab bhattacharjee Aug 28 '17 at 5:40

Hint:

Write $x-5=y$

For $y>0\iff x>5,$ use AM-GM inequality.

What if $y<0?$

$$\left(\frac{x^2}{x-5}\right)'=\frac{2x}{x-5}-\frac{x^2}{(x-5)^2}=\frac{x(x-10)}{(x-5)^2},$$ which gives $x_{min}=10$ and $x_{max}=0$.

Id est, $(10,20)$ it's a minimum point and $(0,0)$ it's a maximum point.

Also we can use the following way.

If $(x,k)$ is an extremum point in our case then the equation $$\frac{x^2}{x-5}=k$$ or $$x^2-kx+5k=0$$ has one real root, which gives $\Delta=0$, which is $k^2-20k=0$ and from here

we can get the answer.

With the techniques from calculus, you can find local minima and maxima using Fermat's Theorem:

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then the derivative at any local extremum is 0.

Thus take the derivative, and set it equal to zero. Doing this gives

$$0 = \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{x^2}{x-5} \right) = \frac{2x(x-5) - x^2}{(x-5)^2} = \frac{x^2 - 10x}{(x-5)^2} \iff x \in \{0, 10\}.$$

Note that $0^2/(0-5) = 0$, and $10^2/(10-5) = 20$. Thus the extreme points are $$(0,0) \qquad\text{and}\qquad (10,20).$$ It can be confirmed that these are a local maximum and minimum, respectively, by applying the second derivative test, or by noting that the first derivative changes sign (from positive to negative, and from negative to positive, respectively) at these points. You can also read this off the graph, if you are willing to trust technology, or check the left- and right-hand limits at the asymptote---the fact that $$\lim_{x\to 5^{-}} \frac{x^2}{x-5} = -\infty$$ and that $$\lim_{x\to -\infty} \frac{x^2}{x-5} = -\infty,$$ combined with the fact that this function is continuous on the interval $(-\infty, 5)$, indicates that the single local extremum in this interval must be a maximum. The other extreme value can be checked by similar reasoning.