A nice calculus problem

What is the least number of critical points of the function f? #shorts

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[SpasmFingers]

I like these, haven't had a math class in year, but calculus is something I don't want to lose

[MrSimpsondennis]

d) I can not determine it from the information given XD

[Macieks]

easy question but with a tricky phrasing

[imnobody]

Two.

FTC tells us that f'(x)=h(x)
For a function to be differentiable, it must be continuous

Therefore

By IVT, there must be at least two because h(x) is continuous and h(2)>0>h(6) and h(9)<0<h(11).

[fackingcopyrights]

i imagined what the function h(x) would look like by the given numbers and came to the conclusion its at least a quadratic function.
since f(x) ist the integral of h(x), f(x) had to be a cubic function and therefore it has to have two critical points.

[EngLhag]

We can see that for every increment of +2 in x, h'(x) consistently (linearly) increases in 4, which means h'(x) is a line with slope equals to 4/2 = 2. So h'(x)=2x+b, and substituting values, we discover that h'(x)=2x-14. By integration, we end with h(x)=x^2-14x+C, and C =39 by comparison and substitution from values. As a 2nd order equation (parabolic), h(x) has by definition 2 roots max. In this case we have the minimum value at h'(7)=0 with h(7)=-10, and the two roots where h(x)=0 are 7+-sqrt(10).

[yllam]

This is equivalent to consider h(x) = 0, there is one root in (2, 6) and one in (9, 11), thetefore f has at least two critical points in (2, 13).

[matthewspence]

Its c, because by the fundamental theorem of calculus, the integral equals h(x) + C. We see h'(x) = 0 only once, meaning it's slope only changes from increasing to decreasing once. Since h(x) has both positive and negative values on either side of h'(7), that means it intersects y=0 twice.

[jaseemk]

Hello sir, can you please tell in which software are you drawing the figures and editing the videos ?

[thewitchking]

This is from an AP AB Calc review textbook my friend literally asked me to solve this one two days ago (BC so we covered this a while back but the teacher explained this concept poorly for his class)

[indigo]

Pretty common calc I problem. Ty for review.

[fatiaibrahim]

The curve intersects the x-axis at two points, but there is actually only one critical point and it is between theese two points.

[XBGamerX]

thats a pretty standard equation in greece. exactly this, no function given at all, just clues.

[MR-repq]

By using Newton's lebinits theorem we get 2 critical points {minimum}

[StuartSimon]

I was taught that the points at which the second derivative is zero or undefined are also critical points because they determine the possible points of inflection. Under that definition, there would be at least three critical points.

[fatiaibrahim]

Because the the h' (x) is a polynomial line (1st degree) with x so h(x) is 2nd degree and it has just one cretical point when h' (x) =0 and finally this graph like polynomial 2nd degree h(x)= ax^2 -bx +c

[royalpoushikan]

Two street lights of height H have a distance L from each other. A person of height h starts from one of them, moving towards the other. Let S be the length of the person’s shadow between the two street lights. There is a maximal value of S when 0<t<L/v and h<H. Find the corresponding values of h and t, respectively.
Can somebody pls solve this sum

[krozjr]

How do we know that h(x) isn’t defined at, for example, x=5? That’d be a critical point, and it wouldn’t mess with the intermediate value theorem, at least the way it’s used here - compare with tan(x), which is differentiable but not at the asymptotes.

[lehtalis]

You can clearly see that with x in [2, 13], h’(x) is linear, which means h(x) is a polynomial of degree 2, f(x) is just integrating h(t), and the critical points of f are the solution of h(x) = 0.

My english is not good so idk if i misunderstood a part of the question but at a basic calculus level thats the trick

[bjyang]

I think the answer is d? Isnt a critical point defined as a point where there exist an open interval in which the values of f(x) is biggest/smallest at that point? So IF a point is a critical point then the derivative is zero(let’s exclude undefined as h (f’) is continuous) however that doesn’t mean all points where the derivative is zero can be critical points as they can be inflection points. so at those points where h is equal to 0, if h’ (which is f”) was also zero then it would be a inflection point and not a critical point . can someone correct me here?

Edit: I confused extreme point and critical point, English isn’t my first language lol