Drawing a function that isn't $x^3$ with a critical number that is not a local extremum. I have a problem that I'm not sure how to do at all. It says,


  Sketch the graph of a function that is continuous on $[1, 5]$ and has an absolute maximum at $2$, an absolute minimum at $5$, and a critical number at $4$, but no local maximum or minimum there.


I know that $f=x^3$ has a critical number that isn't an extremum, but it doesn't have any absolute extrema either.

I am not sure how to draw a critical point that isn't a local minimum or maximum, and yet still has absolute extrema.

How can I draw a function with absolute extrema, and that has a critical point that isn't a local extremum?
Sketch means to draw a picture describing it.  Not to actually find the expression of one.

Assuming continuity and differentiability then:

An absolute maximum at $2$ means: as $1 &lt; 2 &lt; 5$ it must also be a local maximum so the function "plateaus" at two and the $f'(x)=0$ so it "flattens out" at $x = 2$.

A critical point that is neither a local maximum or minimum means (if it is differentiable) means it "flattens" out at $x=4$ but neither "plateaus" or "valleys".  So it "passes through".  

And as 5 is an extreme point of the interval $[1,5]$, an absolute minimum just means the lowest point on the interval. It need not be a critical point.

If you forgive the crappy graphics:
If I have understood your question right then your function should look like this. It should have a Maxima at 2 .minima at 5. It should change its curvature from convex to concave at 4. Notice that these are local Maxima/minima so function can assume values larger/smaller than those at local Maxima/minima