Find the minimum and maximum value of $\frac{x^2-2x+5}{(x-1)^2}$

Question

Find min,max of function $$A = \dfrac{x^2-2x+5}{(x-1)^2}.$$

We see that $A = 1 + \dfrac{4}{(x-1)^2}.$ However I have no next idea. How to solve this problem?

Answer 1

Since when $x\to 1$, $1/(x-1)$ approaches to $\infty$, thus $A$ doesn't have a maximum, or its maximum is $+\infty$. And notice when $x\to \infty$, $1/(x-1)^2\to 0$, it doesn't have a minimum either.

Answer 2

$A \geq 1$ but you can never have $A=1$. So the infimum is $1$ but $1$ is not a minimum. As $x \to 1$ we see that $A \to \infty$ so $A$ is not bounded above. So there is no maximum.

Answer 3

Any real number squared is non-negative: $$(x-1)^2 \ge 0 \implies 0\lt \frac{1}{(x-1)^2} \lt \infty \implies 0\lt \frac{4}{(x-1)^2} \lt \infty \\ \implies 1\lt 1+\frac{4}{(x-1)^2} \lt \infty $$

Answer 4

The derivative is $-8/(x-1)^3$. Since it's never zero it doesn't have a local max/min, then you have to look at the asymptotes(FFjet already done it).

Answer 5

The function $$A= \frac{x^2-2x+5}{(x-1)^2}=1+\frac{4}{(x-1)^2}\quad \forall x\in (-\infty,1)\cup(1,+\infty)$$

Hence, at $x=1$, the function is undefined. The minimum of the function however is when $x=\pm\infty, A=1$ while at $x=1$, the function achieves it maxima since $$\lim_{x\to1}\frac{x^2-2x+5}{(x-1)^2}\to \infty$$ You can also check it through the graph of this function.

It is clear that the function tends towards infinity both from $x\to 1^-, x\to 1^+$