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?
Best Answer
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 $1 < 2 < 5$ 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: