# Is the line really passing through point $a$?

I've answered an exercise on Stewart's Essential Calculus, he asks me to sketch a graph of an example of a function $f$ that satisfies all of the given conditions below:

• $\lim_{x\rightarrow 3^+}f(x)=4$

• $\lim_{x\rightarrow 3^-}f(x)=2$

• $\lim_{x\rightarrow -2}f(x)=2$

• $f(3)=3$

• $f(-2)=1$

The plot below is the right answer. In my answer, the line that is passing through $a$ is passing through point $b$. Is my answer acceptable? I can't feel the guarantee that the line is really passing through point $a$ and I'm thinking that both answers are right.

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How do you get the third condition, $\lim_{x\to-2}f(x)=3$, if your graph goes smoothly through $B=(-2,1)$? – Rahul Oct 25 '12 at 11:41
And if the limit from both sides is $3$, shouldn't it go through the point $P = (-2, 3)$ (instead of $A$ or $B$)? – TMM Oct 25 '12 at 11:58
What is the point $E$ doing there? $f(3)=4\ne3$. – Gerry Myerson Oct 25 '12 at 12:07
I just want to point out, you say "the plot below is the right answer" where it should say the plot below is a right answer, as it may not be unique. So your answer can be good, as long as it respect every conditions. – Jean-Sébastien Oct 26 '12 at 15:11
It may not pass through $a$ but if you pass through $b$ then you won't have condition $3$ respected. The function wants to go to $2$ at $-2$ coming from the left, but it just can't get there as it needs to be $1$ – Jean-Sébastien Oct 26 '12 at 15:21

The slopes and convexity of the function may vary, but the main thing is that the graph of $f$ (what you called 'line') indeed wants to pass through $A$ (in a neighborhood of $x=-2$), but instead, at least at $x=-2$, it is 'sporadicly' valuated according to $B$. Similarly well plotted for $C,D,E$.