Question

1. no calculator let ( f ) be a function defined for all real numbers ( x ). if ( f(x)=\frac{left|4 - x^{2}

ight|}{x - 2} ), then ( f ) is decreasing on the interval (a) ((-infty,2)) (b) ((-infty,infty)) (c) ((-2,4)) (d) ((-2,infty)) (e) ((2,infty))

Explanation:

Step1: Recall the decreasing - function condition

A function $y = f(x)$ is decreasing when $f'(x)<0$.

Step2: Analyze the numerator and denominator of $f'(x)=\frac{|4 - x^{2}|}{x - 2}$

First, factor $4 - x^{2}=(2 + x)(2 - x)$. The absolute - value function $|4 - x^{2}|=

$$\begin{cases}4 - x^{2},&- 2\leqslant x\leqslant2\\x^{2}-4,&x<-2\text{ or }x > 2\end{cases}$$

$.
Since $|4 - x^{2}|\geqslant0$ for all real $x$, we need to find when $\frac{|4 - x^{2}|}{x - 2}<0$. The numerator $|4 - x^{2}|$ is non - negative. For the fraction to be negative, the denominator $x - 2<0$ and $|4 - x^{2}|
eq0$.
The numerator $|4 - x^{2}| = 0$ when $x=\pm2$.
We consider the sign of the fraction $\frac{|4 - x^{2}|}{x - 2}$. When $x<2$, the denominator $x - 2<0$. And $|4 - x^{2}|\geqslant0$.
If $x<2$, then $f'(x)=\frac{|4 - x^{2}|}{x - 2}<0$ (because the numerator $|4 - x^{2}|$ is non - negative and the denominator $x - 2$ is negative).

Answer:

A. $(-\infty,2)$