Question

f(x)= -x^2/(4x+4)

find the first and second derivative, x and y intercept, vertical and horizontal asymptote, x coordinates of the critical points, relative minima and maxima, open intervals where the function is concave up and concave down, when it is increasing and decreasing.

Also sketch the graph of this function

Answer 1

Solution:

$f(x)=y=\dfrac{-x^2}{4x+4}$

$f'(x)=-\dfrac{d}{dx}\left(\dfrac{x^2}{4x+4}\right) \\=-\dfrac{\frac{d}{dx}\left(x^2\right)\left(4x+4\right)-\dfrac{d}{dx}\left(4x+4\right)x^2}{\left(4x+4\right)^2} \\=-\dfrac{2x\left(4x+4\right)-4x^2}{\left(4x+4\right)^2} \\=-\dfrac{x\left(x+2\right)}{4\left(x+1\right)^2}$

Then, $f''(x)=\dfrac{d}{dx}\left(-\dfrac{x\left(x+2\right)}{4\left(x+1\right)^2}\right)$

$=-\dfrac{1}{4}\cdot \dfrac{\frac{d}{dx}\left(x\left(x+2\right)\right)\left(x+1\right)^2-\frac{d}{dx}\left(\left(x+1\right)^2\right)x\left(x+2\right)}{\left(\left(x+1\right)^2\right)^2} \\=-\dfrac{1}{4}\cdot \dfrac{\left(2x+2\right)\left(x+1\right)^2-2\left(x+1\right)x\left(x+2\right)}{\left(\left(x+1\right)^2\right)^2} \\=-\dfrac{1}{2\left(x+1\right)^3}$

For x-intercept, put y=0.

$\dfrac{-x^2}{4x+4}=0 \\ \Rightarrow x=0$

So, x-intercept is (0,0).

For y-intercept, put x=0.

$y=\dfrac{-0^2}{4(0)+4}=\dfrac04=0$

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

For vertical asymptote: put $4x+4=0$

$\Rightarrow x=-1$ is vertical asymptote.

For horizontal asymptotes:

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:

$\lim _{x \rightarrow \infty}\left(-\frac{x^{2}}{4 x+4}\right)=-\infty \\\lim _{x \rightarrow-\infty}\left(-\frac{x^{2}}{4 x+4}\right)=\infty$

Thus, there are no horizontal asymptotes.

SLANT ASYMPTOTES:

Do polynomial long division $-\frac{x^{2}}{4 x+4}=-\frac{x}{4}+\frac{1}{4}-\frac{1}{4 x+4}$

The rational term approaches 0 as the variable approaches infinity.

Thus, the slant asymptote is $y=\frac{1}{4}-\frac{x}{4}.$

So, $\mathrm{Vertical}:\:x=-1,\:\mathrm{Horizontal}:\:y=-\frac{1}{4}x+\frac{1}{4}\mathrm{\:\left(slant\right)}$

For critical points, put f'(x)=0

$-\dfrac{x\left(x+2\right)}{4\left(x+1\right)^2}=0 \\\Rightarrow x=0,x=-2$

x coordinates of critical points are x=0, x=-2

Relative minima and maxima:

Put x=0 in f(x).

f(0)=y=0

Put x=-2 in f(x).

f(-2)=y=1

So, $\mathrm{Minimum}\left(-2,\:1\right),\:\mathrm{Maximum}\left(0,\:0\right)$

Concavity intervals:

Put f''(x)=0

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

No inflections points, but because of denominator, 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 $x=-2$ from $(-\infty,-1)$ and put in f''(x).

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

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

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

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