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I have to sketch a graph that satisfies the conditions:

A. $f(2)=f(4)=0$
B. $f'(x)\lt0$ if $x\lt3$
C. $f'(3)$ does not exist
D. $f'(x) \gt 0$ if $x \gt 3$
E. $f''(x) \lt 0$,$x\ne 3$

I am a bit stuck on how to tell if the second derivative is always negative from a graph? I know that there is a sharp turn at $x=3$, and there is also a minimum there but the second derivative part trips me up.

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    $f''(x)<0$ when the graph is concave down. Are you familiar with that term? – 2017-01-04
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    @pseudoeuclidean Yes, sorry I haven't really been doing calculus that much, so yes and if $f''>0$ its concave up. Thanks. – 2017-01-04
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    Try to plot this one. $$f(x)=\begin{cases}\ln(x-3), \text{ if } x>3\\ \ln(3-x), \text{ if } x<3\end{cases}.$$ – 2017-01-04
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    @Azzo Its a perfect fit. Thanks – 2017-01-04

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It sounds like you understand what you are doing now, but I will answer anyway.

Condition $A$ means that the function has roots at $2$ and $4$.

Condition $B$ means that the function decreases on the interval $(-\infty,3)$.

Condition $C$ means that the function is discontinuous or has a cusp at $x=3$.

Condition $D$ means that the function increases on the interval $(3,\infty)$.

Condition $E$ means that the function is everywhere concave down, except at $x=3$ (which is expected, given the previous conditions).

Notice the symmetry in these conditions about $x=3$. That should help you. Although the problem only asks for a sketch, these conditions are satisfied by the graph of the equation $y=\frac{(x-2)(x-4)}{|x-3|}$, as well as many others.

enter image description here

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    +1 - Though, I do wonder if the function can be *defined* at $x=3$? – 2017-01-04
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    @MarkMcClure Yes, with the equation $y=\sqrt{|x-3|}-1$ – 2017-01-04