The following question is from our reviewer for our upcoming exams:
$\ast$ 3. Sketch the graph of a function $f$ satisfying the following conditions:
(i). $-5,-3,-2,-1,0$ and $2$ are the only zeros of $f$,
(ii). $\displaystyle\lim_{x\to – 3} f(x) = 4; \displaystyle \lim_{x\to-1^{-}}f(x) = -\infty, \displaystyle\lim_{x\to-1^+}f(x) = 0,\displaystyle\lim_{x\to0}f(x) = +\infty,$ and
(iii). $f$ is continuous at all numbers in the open intervals $(-\infty, -3),(-3,-1),(-1,0),(0,+\infty)$.
Here's what I came up with for the past hour:
Am I going the right way? I don't have any clue to what does it mean by "zeroes of f" but what I feel is that it relates to points that are in the $x$-axis that when substituted to the function, it results to zero.
Also, are there some tips/ review links on how can I visualize better on sketching a graph with conditions that satisfies it? I find it really hard to visualize the graph =(
Thank you =)
Best Answer
zeros of $f$ are values $c$ such that $f(c) = 0$.
I believe there is a problem with your graph at $x = -3$.
Your graph is not defined on $(2,\infty)$, so I don't believe it is continuous on $(0,\infty)$
This is what I am getting:
Those should be negative numbers on the left. Hehe.