Sketch a graph with domain −5≤x≤5 where the following hold.

Question

f′(x) >0 On the interval −2<x<3

f′(x) <0 On the interval −5<x<−2 and 3<x<5

f′'(x) >0 On the interval 4<x<5

f′'(x) <0 On the interval −5<x<−2 and -2<x<4

any help would be amazing

Answer 1

Before you draw anything, read the conditions once more around x = −2, because something surprising happens there and it changes what the picture has to look like.

First translate each condition. f' tells you rising or falling, f'' tells you which way the curve bends.

f' > 0 on (−2, 3): f is increasing there.
f' < 0 on (−5, −2) and on (3, 5): f is decreasing on both.
f'' < 0 on (−5, −2) and on (−2, 4): f is concave down on both.
f'' > 0 on (4, 5): f is concave up there.

The surprise at x = −2. Look carefully at what f' is being asked to do on each side. Just left of −2, f' is negative, and since f'' < 0 there, f' is also decreasing, so it is sliding further down into the negatives as you approach −2. Just right of −2, f' is positive, and it is decreasing there too. So f' has to jump from a very negative value straight up to a positive one exactly at x = −2, taking no values in between.

A differentiable function cannot do that. The derivative of a differentiable function always has the intermediate value property, so f' is not allowed to skip over 0. What this forces is that x = −2 has to be a corner, a sharp point where f is not differentiable. That is perfectly legal here, because as the conditions are written the endpoints of the intervals are never included, so nothing at all is being claimed about f' or f'' at x = −2 itself. If you draw a smooth rounded minimum there, your picture will quietly contradict the concavity condition, and spotting that is the real point of this exercise.

Now the shape, reading left to right.

−5 to −2: the curve goes down, and concave down means it keeps getting steeper as it falls. Picture it plunging toward −2, not easing into it.

At −2: a sharp corner, and it is a local minimum, since f falls before it and rises after it. Draw an actual point there, like the bottom of a V, not a smooth U.

−2 to 3: the curve rises, but concave down means the rise keeps flattening, so it comes up steeply out of the corner and gradually levels off.

At 3: it levels out completely and turns over. This is a smooth local maximum, with f' = 0 and f'' < 0.

3 to 4: falling, still concave down, so it steepens as it drops.

At 4: an inflection point. The concavity flips from down to up. The curve is still falling straight through here, it only changes the way it bends.

4 to 5: still falling, but now concave up, so the drop flattens out and the curve starts leveling off toward the right end of the domain.

A fast way to check your drawing. Sweep across your sketch and ask two separate questions about each stretch: is it going up or down, and does it hold water or spill it? Concave up holds water, concave down spills it. Your picture should read: down and spilling, corner, up and spilling, smooth peak at 3, down and spilling, bend at 4, down and holding. If any stretch disagrees with that list, the mistake is in that stretch and not somewhere else.

One last suggestion, since another tutor mentioned building up from f'' to f' to f. That is good general advice, but on this particular problem the faster route is to sketch f' by itself on a separate set of axes first. Once you see f' sitting negative and sliding downward on the left, then leaping up to positive at −2 and sliding down through zero at 3, the corner jumps out at you immediately, instead of being something you only discover after your graph of f already looks wrong.