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Concavity review

Review your knowledge of concavity of functions and how we use differential calculus to analyze it.

What is concavity?

Concavity relates to the rate of change of a function's derivative. A function f is concave up (or upwards) where the derivative f is increasing. This is equivalent to the derivative of f, which is f, being positive. Similarly, f is concave down (or downwards) where the derivative f is decreasing (or equivalently, f is negative).
Graphically, a graph that's concave up has a cup shape, , and a graph that's concave down has a cap shape, .
Want to learn more about concavity and differential calculus? Check out this video.

Practice set 1: Analyzing concavity graphically

Problem 1.1
Select all the intervals where f(x)>0 and f(x)>0.
Choose all answers that apply:

Want to try more problems like this? Check out this exercise.

Practice set 2: Analyzing concavity algebraically

Problem 2.1
f(x)=3x416x3+24x2+48
On which intervals is the graph of f concave down?
Choose 1 answer:

Want to try more problems like this? Check out this exercise.

Want to join the conversation?

  • blobby green style avatar for user Hafsa Mahmood
    if two functions are concave up will their product and sum also be concave up?
    (2 votes)
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    • leaf green style avatar for user kubleeka
      You have two functions f and g, where f''>0 and g''>0.
      If you take the second derivative of f+g, you get f''+g'', which is positive. So their sum is concave up.

      If you take the second derivative of fg, you get the derivative of f'g+fg', or f''g+2f'g'+fg''. f'' and g'' are positive, but the other terms can have any sign, so the whole expression need not be positive.

      For example, consider f(x)=1/x and g(x)=√x³. Both are concave up for x>0, but their product is √x, which is concave down.
      (9 votes)
  • piceratops ultimate style avatar for user Brendan D.
    If f''(x) of a function is never undefined AND is never equal to zero, how can we determine the concavity of the function?
    (0 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      In order for 𝑓(𝑥) to be concave up, in some interval, 𝑓 ''(𝑥) has to be greater than or equal to 0 (i.e. non-negative) for all 𝑥 in that interval.
      The same goes for 𝑓(𝑥) concave down, but then 𝑓 ''(𝑥) is non-positive.

      Saying 𝑓 ''(𝑥) ≠ 0 is not enough to determine the concavity of 𝑓(𝑥), because 𝑓 ''(𝑥) might not be continuous and could thereby change polarity without crossing the 𝑥-axis.
      (6 votes)
  • blobby green style avatar for user ashrafulmd2000
    Is it possible to have an inflection point at x=a for f(x) even if f''(a) does not equal to zero?
    (3 votes)
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  • duskpin ultimate style avatar for user Anna
    Does differentiability at a point matter when determining concavity? For example, if the question asks what interval is the graph of f concave up but the point is not differentiable (it is the junction of a piecewise function), should it be included when giving the intervals?
    (2 votes)
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    • hopper cool style avatar for user JPOgle ✝
      Points to be considered are points where f"(x) = 0 and f"(x) is undefined. When you are finding places where f(x) is concave up or concave down, you are also finding intervals where f'(x) is increasing or decreasing, so we have to consider all critical points of f'(x).
      (2 votes)
  • marcimus pink style avatar for user Emilie Rennie
    what is the definition of concavity in the context of calculus?
    (2 votes)
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  • blobby green style avatar for user 1068362
    just to be clear:
    a function that increases is concave up
    a Rate of Change that increases is concave up
    a graph that DECREASES is concave up
    (1 vote)
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  • aqualine ultimate style avatar for user Liang
    practice set 2 ----> problem 2.2 ----> explanation:

    quote
    g" is undefined for x=-2, therefore, our critical point is x=-2
    unquote.

    x=-2 is called a critical point here too?
    (1 vote)
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    • male robot donald style avatar for user Venkata
      In general, no. Points where f"(x)= 0 are specifically called inflection points. But in the example provided, -2 is also a critical point (Observe that you get x = -2 when you set f'(x) = 0). So, as the only inflection point is also a critical point, they used that term
      (2 votes)
  • blobby green style avatar for user カナンアア ケネス
    Examine the function:
    y=x(x-1)^3
    y'=3x(x-1)^2+(x-1)^3
    Note: I won't finish the question, I'll just ask where do they get (x-1)^3 in y'?
    (1 vote)
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  • blobby green style avatar for user scott albert
    General question on the graphing behavior of f''(x) where f(x) has a vertical and horizontal asymptote: I'm working with f(x)=(x^2)/(x+1). The graph of f'(x) makes sense as it (they?) crosses the x axis at x=-2 and x=0, where the slope of f(x)'s 2 parabolas = 0. But I don't understand the f''(x) graph. Both f(x) parabolas have negative and positive slopes, yet f''(x) never crosses the x axis. How do I figure this out? Thank you!
    (1 vote)
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  • orange juice squid orange style avatar for user Zy
    Ok, what really confuses me is saying that the concave up graph of f is increasing when it clearly looks that the tangent lines of the graph are decreasing, or negative, until the minimum value, likewise if f is concave down and the tangent lines look positive until the maximum value. Are we speaking in terms of f' graph for f where it shows this?
    (1 vote)
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    • blobby green style avatar for user Beth S.
      For the concave - up example, even though the slope of the tangent line is negative on the downslope of the concavity as it approaches the relative minimum, the slope of the tangent line f’(x) is becoming less negative... in other words, the slope of the tangent line is increasing. so over that interval, f”(x) >0 because the second derivative describes how the slope of the tangent line to the function is changing at any given x. Over this interval, we can see the slope of the tangent to the function becomes less and less steep, and less and less negative, until it reaches the minimum point, where f’(x) =0.
      (1 vote)