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Problem 33 Easy Difficulty
Suppose $ f $ is a continuous function where $ f(x) > 0 $ for all $ x $, $ f(0) = 4 $, $ f'(x) > 0 $ if $ x < 0 $ or $ x > 2 $, $ f'(x) < 0 $ if $ 0 < x < 2 $, $ f"(-1) = f"(1) = 0 $, $ f"(x) > 0 $ if $ x < -1 $ or $ x > 1 $, $ f"(x) < 0 $ if $ -1 < x < 1 $.
(a) Can $ f $ have an absolute maximum? If so, sketch a possible graph of $ f $. If not, explain why?
(b) Can $ f $ have an absolute minimum? If so, sketch a possible graph of $ f $. If not, explain why?
(c) Sketch a possible graph for $ f $ that does not achieve an absolute minimum.
Answer
(a) Intuitively, since $f$ is continuous, increasing, and concave upward for $x>2,$ it cannot have an absolute maximum. For a
proof, we appeal to the MVT. Let $x=d>2 .$ Then by the MVI, $f(d)-f(2)=f^{\prime}(c)(d-2)$ for some $c$ such that
$2<c<d .$ So $f(d)=f(2)+f^{\prime}(c)(d-2)$ where $f(2)$ is positive since $f(x)>0$ for all $x$ and $f^{\prime}(c)$ is positive since $f^{\prime}(x)>0$ for $x>2 .$ Thus, as $d \rightarrow \infty, f(d) \rightarrow \infty,$ and no absolute maximum exists.
(b) Yes, the local minimum at $x=2$ can be an absolute minimum.
(c) Here $f(x) \rightarrow 0$ as $x \rightarrow-\infty,$ but $f$ does not achieve an absolute minimum.
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Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
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Video Transcript
s O. We're told that after the continued function where F X is greater than zero for all acts and were pulled a set of different condition and we're being asked questions about thie graph that is produced from this condition which have drug right here. And we're told Ken Hat can f have an absolute maximum. The question is, the answer to that would be No, because even without drawing the graph, we know that the function is always positive and we also know it is increasing from negative info for Ex Lesson Joe, so absurd to negative infinity and for all expected them to from two to positive infinity that I missed the functional always keep increasing. So there's so there's all will always be one point that is greater, then the next and the next and the next. So it is not possible. So no, Herbie can have haven't absolute minimum. If so, sketchy Ponce photograph. This could be true depending on the conditions. So in my graph, uh, that's a prissy. This is a case in which, after not achieve in absolute men, this will go often to go under, too. In another case in such a graph that will. So these two are possible possible? So for be such a graph would look something like this. Ah, the straight line. And then I'm sure the rough guess so Look, something like there will be a sort of ass in tow. Oh, that is horrible. Ah, what are they like? They're dip really well like that. And this would be a the absolute men, and the most likely occurrence would occur in X equals two.
Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Anna Marie V.
Campbell University
Michael J.
Idaho State University
Joseph L.
Boston College
