Question

f(x)= 3/(x-1)

find the x and y intercepts, vertical and horizontal asymptotes, x coordinates of the critical points, open intervals where the function is concave up and concave down, relative minima and maxima.

Could you also sketch the graph of the function as well.

Answer 1

Solution:

$f(x)=y=\dfrac3{x-1}$

For x-intercept, put y=0.

$0=\dfrac3{x-1} \\\Rightarrow x-1=\dfrac30$

x-intercept does not exist.

For y-intercept, put x=0.

$y=\dfrac3{0-1} \\ \Rightarrow y=-3$

So, y-intercept is (0,-3).

For vertical asymptotes: Put $x-1=0$

So, vertical asymptote: $x=1$

Line y=L is a horizontal asymptote of the function y=f(x), if either $\lim _{x \rightarrow \infty} f(x)=L$ or $\lim _{x \rightarrow-\infty} f(x)=L$ , and L is finite.

Calculate the limits:

$\lim _{x \rightarrow \infty}\left(\frac{3}{x-1}\right)=0 \\ \lim _{x \rightarrow-\infty}\left(\frac{3}{x-1}\right)=0$

Thus, the horizontal asymptote is y=0.

Horizontal asymptote: $y=0$

Next, $f(x)=\dfrac3{x-1}$

$f'(x)=\dfrac{d}{dx}\left(\dfrac{3}{\left(x-1\right)}\right) \\=3\dfrac{d}{dx}\left(\left(x-1\right)^{-1}\right) \\=-\dfrac{3}{\left(x-1\right)^2}$

put f'(x)=0

$-\dfrac{3}{\left(x-1\right)^2}=0$

No x-coordinate exists for critical values.

So, extreme points exist either.

Now, $f'(x)=-\dfrac{3}{\left(x-1\right)^2}$

$f''(x)=-3\dfrac{d}{dx}\left(\dfrac{1}{\left(x-1\right)^2}\right) \\=-3\dfrac{d}{dx}\left(\left(x-1\right)^{-2}\right) \\=\dfrac{6}{\left(x-1\right)^3}$

Intervals are $(-\infty,1), (1,\infty)$.

\mathrm{If}\:f\:''\left(x\right)>0\:\mathrm{then}\:f\left(x\right)\:\mathrm{concave\:upwards.} \\ \mathrm{If}\:f\:''\left(x\right)<0\:\mathrm{then}\:f\left(x\right)\:\mathrm{concave\:downwards.}

Take $x=0$ from $(-\infty,1)$ and put in f''(x).

We get $f''(x)<0$, so concave downward.

Take $x=2$ from $(1,\infty)$ and put in f''(x).

We get $f''(x)>0$, so concave upward.

$\mathrm{Concave\:Downward}:-\infty \:<x<1,\:\mathrm{Concave\:Upward}:1<x<\infty$