What is the range of the expression $5/(x^2+|2x-4|)$?

Question

What is the range of the expression $5/(x^2+|2x-4|)$?

I did use a graphing calculator to find that the range is $0< y \le 5/3$, but how would I find this by hand? The absolute value sign gets in the way of my efforts.

Answer 1

For $x \ge 2$, the denominator is $x^2 + 2x - 4 \ge 2^2 + 2 \cdot 2 - 4 = 4$.

For $x < 2$, the denominator is $x^2 - 2x + 4 = (x - 1)^2 + 3$. This is minimized at $x = 1$, and can attain any value in the interval $[3, \infty)$.

Hence, the range of the denominator is $[3, \infty)$, and the range of the function is $(0, 5/3]$.

Answer 2

You basically want to find when the denominator $z=x^2+|2x-4|$ will be max and min. Consider the two intervals: $$1)\begin{cases}x<2 \\ z=x^2-2x+4=(x-1)^2+3\end{cases} \Rightarrow \begin{cases} z(1)=3 \ \ \text{min} \\ z(-\infty)=+\infty \ \ \text{no max}\end{cases} \Rightarrow 0<\frac{5}{z}\le \frac53. \\ \text{OR}$$ $$2)\begin{cases}x\ge 2 \\ z=x^2+2x-4=(x+1)^2-5\end{cases} \Rightarrow \begin{cases} z(2)=4 \ \ \text{min} \\ z(+\infty)=+\infty \ \ \text{no max}\end{cases} \Rightarrow 0<\frac5z\le \frac54.$$ Thus, the range of the function is the bigger interval $(0,\frac53\big{]}$.