Question

The range of the values of $\frac{x}{x^2+4}$ for all real value of $x$ is:

• A. $\frac{-1}{4}\le y\le \frac{1}{4}$
• B. $\frac{-1}{3}\le y\le \frac{1}{3}$
• C. $\frac{-1}{2}\le y\le \frac{1}{2}$
• D. none of these

Answer 1

The correct option is A $\frac{-1}{4}\le y\le \frac{1}{4}$

$f\left(x\right)=\frac{x}{x^2+4}$

$f^{\prime }\left(x\right)=\frac{x^2-4}{(x^2+4{)}^2}$

putting $f^{\prime }\left(x\right)=0$ we get

$\frac{x^2-4}{(x^2+4{)}^2}=0\Rightarrow x^2-4=0$

$x=2$ or $x=-2$

Now,

$f"\left(x\right)=\frac{(x^2-4)\left[2\right(n^2+4\left)\right(2n\left)\right]-(x^2+4)\left(2n\right)}{(x^2+4{)}^4}$

Putting $x=2$ in $f"\left(x\right)$, we get $f"\left(x\right)<0$ and, by putting $x=-2$ in $f"\left(x\right)$ , we get $f"\left(x\right)>0$

So,

$f\left(2\right)$ is the maximum and $f(-2)$ is the minimum.

$f\left(2\right)=\frac{2}{2^2+4}=\frac{2}{8}=\frac{1}{4}$

$f(-2)=\frac{-2}{2^2+4}=\frac{-2}{8}=\frac{-1}{4}$

$\therefore \frac{-1}{4}\le f\left(x\right)\le \frac{1}{4}$