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Worked example: Approximation with local linearity

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.F (LO)
,
CHA‑3.F.1 (EK)
Finding the equation of a tangent line at a point of a curve by knowing the derivative at that point. Then using that equation to approximate the value of the function at close by x-values.

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  • piceratops seed style avatar for user Billy Gear
    Why did Sal change to the point-slope form for this question after showing us the equation L(x) = f(x)+f'(x)(x-a) in the Local Linearity video? I am confused as to when to use that equation now.
    (11 votes)
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    • blobby green style avatar for user Sridhar S Menon
      Actually, the format used here was based on the fundamental definition of a slope. That is, the slope of a line is the change in y quantity over the change in x quantity. 4 is the slope here and (y-1) is the change in y quantity while (x-2) is the change in x quantity
      (1 vote)
  • spunky sam blue style avatar for user Mircea Nemteanu
    Am I wrong to say that maybe we should use the extra information and the result may be different? The reasoning I use (right or wrong?) is that using the second derivative we could get an approximation for the first derivative at the point 1.9 then we can use the average of the calculated approximation of the 1st derivative at x=1.9 and the given 1st derivative value at x=2 and that would give us a more precise approximation of f(1.9). I happens that the better approximation of f(1.9) is about 0.615 (if I calculated it right), which would still make 0.6 the best answer, but it could have been different.
    (4 votes)
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  • blobby green style avatar for user Harikesh
    : You can follow this if feeling lost with point slope form.

    You can look at it in this way.

    General equation of line is y = mx + b, where m = slope of the line and b = Y intercept.

    We know that f(2) = 1 i.e. line passes through (2,1) and we also know that slope of the line is is 4 because derivative at x = 2 is 4 i.e. f'(2)= 4. Hence we can say that

    y = mx + b
    1 = 4*2 + b
    1 - 8 = b
    b = -7.

    Now we have equation of line : y = 4x - 7
    We can substitute x = 1.9 to find it's y value.

    y = 4 * 1.9 - 7
    y = 0.6

    Hope that helps.
    (2 votes)
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  • starky ultimate style avatar for user DrewMTR
    Is there a way we can make use of the 2nd derivative to make a better approximation of the value of f(1.9)? I know the question is only asking via local linearity, but is it possible to extend that intuition using higher order derivatives?
    (2 votes)
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    • hopper cool style avatar for user Iron Programming
      You are hitting on an interesting point.
      There is a very interesting concept in Single Variable calculus called the Taylor Series. This is basically a technique used to approximate a function at f(c) based upon its derivatives.

      I won't say much here, but if you are interested then keep working hard until you get to that point in your calculus curriculum. :-)
      (3 votes)
  • aqualine ultimate style avatar for user Martin
    What if the question then asks whether the value the approximation is greater or less than the actual value of the function? How do we find this?
    (1 vote)
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    • leaf grey style avatar for user Alex
      Depends on concavity. It's an overestimation if it's concave down, underestimation if concave up, exact value if zero concavity (linear), and needs more information if concavity changes.
      (2 votes)
  • male robot hal style avatar for user Jimmy
    Question about the 2nd derivative is needless in this problem.
    A tangent line can be above or below the function, right? I don't know the technical term here, but intuitively, concavity feels right, concave up or down, right?
    In this video, Sal assumed the actual function is to the left of the tangent, but we don't know that, right? What if the function is to the right of the tangent. So, my question is, does the 2nd derivative really useless in terms of figuring out whether the function is opening up-left or down-right? Thanks.
    (1 vote)
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    • male robot hal style avatar for user Jimmy
      I was trying to think about, in this problem, is f'' really UNNEEDED. After given it a bit more thought, I think in this problem, it is unneeded. Because, original, I was thinking we know the point on the curve is (2, 1), we know the slope at that point is 4, and the also the 2nd derivative is 3. Now, I think in this case, we actually know the function is concave up, but that info doesn't really help approximate f(1.9), the only thing we can do is use the 1st derivative to approximate the answer. But, there is a vague idea in the back of my mind saying, we can then use the 2nd derivative f''(2) = 3, to infer the actual output of the f(1.9) must in between our approximated value(0.6) and 1.
      Moreover, think about it, what if the question says f''=-3. That doesn't change our approximation, but there's something different implied, maybe f(1.9) is less than 0.6.
      Anyway, this all my intuition, there's a great chance what I said is nonsensical.
      (1 vote)
  • duskpin ultimate style avatar for user Coleton O'Donnell
    Something of note, it would also be possible to use the equation:
    f(x) + f'(x)(a-x)

    Where x is equal to our point on the curve (in this video x = 2) and a is equal to what we are trying to approximate (in this case 1.9):

    f(2) + f'(2)(1.9 - 2)
    1 + 4(-0.1)
    1 - 0.4
    0.6
    (1 vote)
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  • leaf orange style avatar for user Mario S
    A little help

    λ=(
    f

    (c)
    f(5102)−f(2015)

    )(
    f
    2
    (c)
    f
    2
    (2015)+f
    2
    (5102)+f(2015)f(5102)

    )
    Let f:[2015,5102] \rightarrow [0,\infty)f:[2015,5102]→[0,∞)be any continuous and differentiable function. Find the value of \lambdaλ, such that there exists some c\in [2015,5102]c∈[2015,5102] which satisfies the equation above.
    (0 votes)
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Video transcript

- [Instructor] We're told the function f is twice differentiable with f of two equals one, f prime of two is equal to four, and f prime prime of two is equal to three. What is the value of the approximation of f of 1.9 using the line tangent to the graph of f at x equals two? So pause this video and see if you can figure this out. This is an actual question from a past AP Calculus exam. All right now let's do this together. And if I was actually doing this on an exam, I would just cut to the chase and I would figure out the equation of the tangent line at x equals to go through the point two comma one, and then I would figure out okay, when x is equal to 1.9, what is the value of y? And that would be my approximation. But for the sake of learning and getting the intuition here, let's just make sure we understand what's happening. So let me graph this. So let's say that's my y-axis, and then this is my x-axis. And this is x equals one. This is x equals two. This is y equals one. We know that the point two comma one is on the graph of y is equal to f of x. So we know that point right over there is there. And we also know the slope of the tangent line. The slope of the tangent line is four. So it's gonna look something like this. It's gonna probably even be a little steeper than that. The slope of the tangent line is gonna look something like that. And we don't know much more about it. We know the second derivative here. But what they're asking us to do is without knowing what the function actually looks like, the function might look something like this. Let me just draw something. So the function might look, might look something like this. We're trying to figure out what f of 1.9 is. So if x is 1.9, f of 1.9, if that's the way the function actually looked, might be this value right over here. But we don't know for sure because we don't know much more about the function. But what they're suggesting for us to do is use this tangent line. If we know the equation of this tangent line here, we could say well what does that tangent line equal when x equals 1.9? When x equals 1.9, it equals that point right over there. And then we could use that as our approximation for f of 1.9. Well to do that we need to know the equation of the tangent line. And we could do that in point-slope form. We would just have to say y minus the y value that we know is on that line, the point two comma one we know is on that line, so y minus one is going to be equal to the slope of our tangent line, which we know is going to be equal to four, times x minus the x value that corresponds to that y value, so x minus two. So now we just have to substitute x equals 1.9 to get our approximation for f of 1.9. So we would say y minus one is equal to four times 1.9 minus two. 1.9 minus two is negative 0.1. And let's see four times negative 0.1, this all simplifies to negative 0.4. Now you add one to both sides, you get y is equal to, if you add one here you're gonna get 0.6. So this, I didn't draw it quite to scale, 0.6 might be something closer to right around there, but there you go. That is our approximation for f of 1.9, which is choice B. And we're done. And one interesting thing to notice, we didn't have to use all the information they gave us. We did not have to use this information about the second derivative in order to solve the problem. So if you ever find yourself in that situation, don't doubt yourself too much. Because they will sometimes give you unneeded information.