Question

For $x\in (1,\infty )$, the graph of the following function is:

$y=\frac{(x+3)}{(x-1)}$

• A. Constant
• B. Monotonically Increasing
• C. Monotonically Decreasing
• D. None of These

Answer 1

The correct option is C Monotonically Decreasing

$y=0$ implies $\frac{x+3}{x-1}=0$
$\Rightarrow x=-3$
Hence the graph of $y=f\left(x\right)$ cuts the x axis at $x=-3$.
Now as $x\to 1$, $y\to \infty$.
Also $y=f\left(x\right)$ cuts the y axis at $(0,-3)$.
$y^{\prime }=\frac{x-1-(x+3)}{(x-1{)}^2}$
$=\frac{-4}{(x-1{)}^2}$

Now, $(x-1{)}^2>0$ $\forall x\in R$
Therefore, for $x\in (1,\infty )$, $y^{\prime }<0$
Thus, $y=f\left(x\right)$ is decreasing function for all real values of $x$.

Hence, $y=f\left(x\right)$ is monotonically decreasing function for $x\in (1,\infty )$