• Mar 19th 2006, 05:31 PM
Vigo
Let f be the function given by f(x)=2x/sqrt(x^2+x+1)

b)Write an equation for each horizontal asymptote of the graph f.

Note: f'(x)=(x+2)/(x^2 +x+1)^.5

• Mar 19th 2006, 07:43 PM
ThePerfectHacker
Quote:

Originally Posted by Vigo
Let f be the function given by f(x)=2x/sqrt(x^2+x+1)

The fraction is underfined when the denominator is zero. Also, when the radical is negative. Thus, $x^2+x+1=0$ or $x^2+x+1<0$. Thus, $x^2+x+1>0$ is the domain. But $x^2+x+1$ is always positive, because the leading coefficient is positive and its discrimanat is negative. Thus, this function is defined for all real numbers.
Quote:

Originally Posted by Vigo
b)Write an equation for each horizontal asymptote of the graph f.

The horizontal tangents is when the derivative approaches infinity. Since is derivative is,
$f'(x)=\frac{x+2}{\sqrt{x^2+x+1}}$ that happens when the denominator is zero. Thus, $\sqrt{x^2+x+1}=0$ thus, $x^2+x+1$ but that never happens because we said that $x^2+x+1$ is always a positive number. Thus, it has no horizontal tangents.
• Mar 19th 2006, 09:47 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
The horizontal tangents is when the derivative approaches infinity. Since is derivative is,
$f'(x)=\frac{x+2}{\sqrt{x^2+x+1}}$ that happens when the denominator is zero. Thus, $\sqrt{x^2+x+1}=0$ thus, $x^2+x+1$ but that never happens because we said that $x^2+x+1$ is always a positive number. Thus, it has no horizontal tangents.

No a horizontal tangent is when the derivative approaches zero.

$
\frac{d}{dx}\frac{2x}{\sqrt{x^2+x+1}}=\frac{x+2}{( x^2 + x + 1)^{3/2}}$

There is a zero of this at $x=-2$, but that is not an asymtote.
Also the derivative goes to zero as $x \to \pm \infty$, so if
the limits of $f(x)$ as $x \to \pm \infty$ are finite these
will give horizontal asymtotes.

Now:

$
f(x)=\frac{2x}{\sqrt{x^2+x+1}}=\frac{2\ \mbox{sign}(x)}{\sqrt{1+1/x+1/x^2}}
$

So as $x \to \infty\ \ f(x) \to 2$, and as $x \to -\infty\ \ f(x) \to -2$, giving horizontal asymtotes at $y=\pm 2$

RonL
• Mar 23rd 2006, 05:52 AM
Vigo
I see how you got the equation for the horizontal asympototes but when I graph the function, y=-2 does not quite fit the horizontal asymptote on the bottom. Why is that so?
• Mar 23rd 2006, 06:19 AM
CaptainBlack
Quote:

Originally Posted by Vigo
I see how you got the equation for the horizontal asympototes but when I graph the function, y=-2 does not quite fit the horizontal asymptote on the bottom. Why is that so?

Probably because you have not extended the plot far enough to the left.
Here is a little table which shows what is going on:

$
\begin{array}{cc}x&f(x)\\-1&-2\\-2&-2.31\\-3&-2.27\\-4&-2.22\\-5&-2.18\\-10&-2.10\\-100&-2.01 \end{array}
$

RonL