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## Approximating values of a function using local linearity and linearization

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# Worked example: Approximation with local linearity

AP Calc: CHA‑3 (EU), CHA‑3.F (LO), CHA‑3.F.1 (EK)

## 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.

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