Can anyone please tell me how to tell that the following graph is $\frac{1}{x^2 - 4}$ ?
My Attempt : From the graph , we can see $\lim_{x \to 2+} = \infty = \lim_{x \to -2-} $ and $\lim_{x \to 2-} = -\infty = \lim_{x \to -2+} $ .
But I can not conclude from these that the graph will be $ \frac{1}{x^2 - 4}$.
Can anyone please tell me how to show that the graph will be $ \frac{1}{x^2 - 4}$.
On
I think in addition to the limits to infinity that @PNDas gave above, you need to state the table of values for selected points on the graph. You can show the graph with its limits, limits to infinity and table of values at important points for x.
On
First is an even function. Next it is discontinnuous at $x=\pm 2$. Check $f(1.9)$ is large negative and $f(2.1)$ is large positive. $f'(x)=-(x^2-4)^{-2} -2x$, so $f'(0)=0$ and the first derivative changes sign when we pass from say $x= -0.1$ to $x=0.1$ from positive to negative, hence there is local max at $x=0$ and $f_{max}=f(0)=-1/4$. Lastly, $f(\pm \infty)=0$. No wonder it looks like as it does.
On
Roadmap:
0)$f$ is symmetric about $y-$axis (Why?)
Consider $f$ for $x \ge 0$, $x \not=2$.
1)Interval $[0,2);$ $f(0)=0;$
What happens for $x \rightarrow 2^{-}$?
4)Interval $(2,\infty);$
What happens for $x \rightarrow 2^{+}$?
5)Consider $x \rightarrow \infty.$
6)Reflect the graph about $y-$axis.
This is the plot of
$$\frac{x^2+10}{10(x^2-4)}.$$
It is virtually impossible to distinguish it from the given one.