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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 4

Lesson 6: Approximating values of a function using local linearity and linearization

# Worked example: Approximation with local linearity

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.

## Want to join the conversation?

• : 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.
• I find this a much better approach and more consistent with what we learned earlier about the slope/line formula.
• 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.
• 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)
• 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.
• The method you used is perfectly legitimate and would likely give a better approximation of 𝑓(1.9).

Just keep in mind that it's not the method the questioner wants us to use.
• 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?
• 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. :-)
• 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
• 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)
• 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.
• 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)
• 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)
• 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.