Tangent line approximation can also be called local linearization, linear approximation, and probably a bunch of other names. The important thing is that you're using a line to approximate a curve.
Here's a reminder of how tangent line approximation works:
Suppose we know the value of f at a particular value of x:
and we want to know the value of f at a nearby value of x:
We draw the tangent line to f at the point we know:
Then we find the value on the tangent line at the nearby value of x:
This tangent line value is close to the value we actually wanted.
Using the language from the last section, let's talk this through again.
We know yold:
We want to know the value of f if we change x by Δ x:
We draw the tangent line to f at the point we know:
and we use the tangent line to find ynew, which is close to the value we actually wanted.
Sample Problem
Another kind of problem you may run into is the kind where you're given a formula for the function f and asked to use a tangent line to approximate f at some particular value x*.
To do this, pick some x = a close to x*. Choose a so that f (a) and f '(a) are easy to calculate. Then we have yold = f (a), Δ x = x* - a, and slope = f '(a).
From there you know how to calculate ynew, which is the approximation you want.
Practice:
Let f (x) = 4x2 + 3. (a) Find the tangent line to f at x = 3.
(b) Use the tangent line from part (a) to approximate f at 3.001. | |
(a) To find a tangent line we need a point and a slope. The point we want is (3,f (3)) = (3,4(3)2 + 3) = (3,39). The slope is f '(3) = 8(3) = 24. Putting this together (link to point-slope form), we get the line y = 24x - 33. (b) We put x = 3.001 into the tangent line and see what we get: 24(3.001) - 33 = 72.024-33 = 29.024. | |
If f (1) = 2 and f ' (1) = 3, estimate f (1.2). | |
We're told the value and slope of f at 1 and asked to estimate f at 1.2. This means we want to draw the tangent line to f at x = 1, and find the value of that tangent line when x = 1.2. In other words, we have yold = 2 , slope = 3, and Δ x = 0.2. We're asked to find ynew, which we know how to do. 
We never had to find the equation for the tangent line! It was enough to know the slope of the tangent line, the value of y we started at, and the amount by which x changed. | |
The picture below shows the tangent line to the function f at x = 0. Estimate f (-0.1). 
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We have yold = -2 and Δ x = -.1 (yes, Δ x is negative). We want ynew, which is the value of the tangent line when x = -0.1. In order to find ynew we need to know the slope of the line. Since we're given two points on the line, we can figure that out. The slope of the tangent line is 
Then 
When doing these problems, so long as Δ x is small we expect yold and ynew to be pretty close together. After all, yold and ynew are both on the same line, and if Δ x is small we're not moving very much on the line. In the previous example, we had yold = -2 and ynew = -2.2. These are pretty close together, so the value ynew = -2.2 is believable. An answer like ynew = 20 would be a call to double-check our calculations. The values ynew, yold, and Δ x can also be used to find the slope of a tangent line to f. Don't think too hard about this! We're just revisiting the formula 
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The picture below shows a tangent line to f at x = a. What is f '(a)? 
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We don't even need to worry about which one is ynew and which one is yold. We just look at the two points and find rise and run. The slope of the line is 
Since this line is tangent to f at a, the slope of the line must agree with the slope of f at a. We conclude 
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Let f (x) = cos2 x. Use a tangent line approximation to estimate f (0.1). | |
We need to decide where to draw the tangent line. Since .1 is close to 0, x = 0 is a good bet. Then yold = f (0) = 1 and Δ x = 0.1 The slope of the tangent line is f '(0). First let's find the derivative of f: f '(x) = 2cos x(-sin x) so f '(0) = 0. We get 
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Let f (x) = x5 + 3x4 - 2x2 - x. Use a tangent line approximation to estimate f (.9). | |
Let's draw the tangent line at x = 1, since 1 is a convenient number that's close to .9. Then we have Δ x = -.1 yold = f (1) = 1 + 3 - 2 - 1 = 1 To find the slope we need to know f ': f'(x) = 5x4 + 12x3 - 4x - 1 The slope is f '(1) = 5 + 12 - 4 - 1 = 12. This is enough to find the approximation: 
It's kind of weird that we got a negative number, but if we look at a graph of the function and tangent line together we can see why: the tangent line drops below the x-axis while the function is still above the x-axis. You know how to do tangent line approximations. You know how to do tangent line approximations by finding just the slope of the tangent line rather than bothering with the whole tangent line equation. The only thing we've changed for these problems is that we're no longer telling you where to draw the tangent line. Instead, you get to figure that out for yourself. To pick the right spot for the tangent line, take x = a near where you're asked to estimate f, and make sure f (a) and f '(a) are easy to compute. | |
Can a tangent line approximation ever produce the exact value of the function? Why or why not? | |
Yes. If the function f is a straight line then the tangent line at any point will be the same as the function. This means the tangent line approximation will produce the same value as the function. | |
The function y = f (x) is a solution to the IVP 
Approximate f (1.01). | |
We know f (1) = 2 and we want to estimate f (1.01). Sounds like Δ x = 0.01 and yold = 2. The differential equation tells us the derivative of y, so we can use the differential equation to figure out the slope of the tangent line. The slope of y at (1,2) is 
So 
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For the function,
(a) Find the tangent line to the function at the specified value of x.
(b) Use the tangent line from (a) to estimate the value of the function at (x + Δ x).
f (x) = x3 + 2x + 1, x = 1, Δ x = .2
Answer
(a) Let's find the tangent line. First we find the value of f when x = 1:
f (1) = 13 + 2(1) + 1 = 4
so our point is (1,4). Now we find the slope. Since f'(x) = 3x2 + 2, the slope at the point (1,4) is
f ' (1) = 3(1)2 + 2 = 5.
Putting this together, the tangent line is
y = 5x - 1.
For a sanity check, we can put this into our calculator and make sure it looks like a tangent line, which it does.
(b) We want to estimate f (1 + .2), so we put x = 1.2 into the equation for the tangent line:
5(1.2) - 1 = 6 - 1 = 5
For the function,
(a) Find the tangent line to the function at the specified value of x.
(b) Use the tangent line from (a) to estimate the value of the function at (x + Δ x).
Answer
(a) The function value at 
is
The derivative function is
f '(x) = 2sin xcos x
so the slope of our tangent line is
Putting this together, the tangent line is
For a sanity check, we can put this into our calculator and make sure it looks like a tangent line, which it does.
(b) To estimate 
, we plug this value into the tangent line:
For the function,
(a) Find the tangent line to the function at the specified value of x.
(b) Use the tangent line from (a) to estimate the value of the function at (x + Δ x).
f (x) = log2 x, x = 8, Δ x = .01
Answer
(a) When x = 8 we have
f (8) = log2 8 = 3.
The derivative of f is
so the slope of the tangent line is
This means the equation for the tangent line is
(b) We plug 8.01 into the tangent line equation:
For the function,
(a) Find the tangent line to the function at the specified value of x.
(b) Use the tangent line from (a) to estimate the value of the function at (x + Δ x).
f (x) = 4x + ex, x = 0, Δ x = 0.05
Answer
(a) We know f (0) = 4(0) + e0 = 1, so the point we need for the tangent line is (0,1). The derivative of f is 4 + ex, so the slope of the tangent line is
f '(0) = 4 + e0 = 5.
Putting this together, the equation for the tangent line is
y = 5x + 1
(b) Plug 0 + 0.05 = 0.05 into the tangent line equation:
5(0.05) + 1 = 1.25
For the function,
(a) Find the tangent line to the function at the specified value of x.
(b) Use the tangent line from (a) to estimate the value of the function at (x + Δ x).
Answer
(a) The slope of the tangent line at 
is
The value of the function at 
is 1, so we get the following tangent line:
(b) We plug
into the tangent line and get
It turns out that when you know a value of f at a particular x and you're asked to approximate the value of f somewhere nearby, you don't need to go to the trouble of finding the tangent line equation. As we saw earlier (link back up to where we reviewed slope), it's enough to know the slope of the tangent line, yold, and Δ x. That's enough information to find ynew.
If f (3) = 4.2 and f '(3) = 0.15, estimate f (3.1).
Answer
We have yold = 4.2, Δ x = 0.1, and slope = 0.15. So
If 
and 
, estimate 
.
Answer
We have 
, slope = 
, and 
. So
This is believable. We expect yold and ynew to be close together, and they are:
The picture below shows the tangent line to f at -1. Estimate f (-.9).
Answer
We have yold = -1 and 
We need to know f '(-1), also known as the slope of the tangent line. Since we have two points on the line we can find this.
The slope of the line is
Now we can find
This is close to yold = -1, so our answer makes sense.
The picture below shows the tangent line to f at x = 3. Find f '(3).
Answer
We know that f '(3) is the same as the slope of the tangent line to f at x = 3, so all we need to do is find the slope of this line.
We get
If f (2) = 7 and a tangent line approximation at x = 2 estimates f (2.1) to be 7.2, what is f '(2)?
Answer
We have f (2) = 7. A tangent line approximation at x = 2 estimates f (2.1) ≅ 7.2. This means if we draw the tangent line to f at 2, the point (2.1,7.2) is on that line.
We know that f '(2) is the same as the slope of the tangent line to f at x = 2, so we need to find the slope of this tangent line. The slope is
so we conclude
f '(2) = 2.
Let f (x) = 3x2 - 4x + 1. Use a tangent line approximation to estimate f (2.01).
Answer
Let's take the tangent line approximation at x = 2, since that's a nice round number that's close to 2.01. Then
Δ x = 0.01
yold = 3(2)2 - 4(2) + 1 = 5.
Since f '(x) = 6x - 4, the slope of the tangent line at x = 2 is
f '(2) = 6(2) - 4 = 8.
therefore
Let f (x) = 7x2 - 3x + 4. Use a tangent line approximation to estimate f (2.99).
Answer
Since 2.99 is close to 3, we'll take the tangent line to f at 3. Then Δ x = -0.01 and
yold = f (3) = 7(9) - 3(3) + 4 = 58.
The derivative of f is
f '(x) = 14x - 3
so the slope of the tangent line at 3 is
f '(3) = 14(3) - 3 = 39.
We can now estimate f (2.99) by finding ynew on the tangent line:
Let f (x) = ex. Use a tangent line approximation to estimate f (0.05).
Answer
Let's take the tangent line at x = 0, since that's close to 0.05. Then Δ x = 0.05.
Since f '(x) = f (x) = ex, both yold and the slope of the tangent line equal e0 = 1.
Then
Let f (x) = 2x. Use a tangent line approximation to estimate f (3.02).
Hint
Wait until the end to round off your answer
Answer
Let's take the tangent line to f at x = 3, which is close to .02. Then Δ x = .02. We have
yold = f (3) = 23 = 8.
The derivative of f is
f '(x) = 2x ln 2
so the slope of the tangent line at 3 is
f '(3) = 8 ln 2.
Now we can do the estimation:
We didn't round until the very last step.
Let f (x) = sin (5x). Use a tangent line approximation to estimate f (3).
Answer
Following the hint, let's draw the tangent line at x = π, which is close to 3. Then
Δ x = 3 - π.
If we're drawing a tangent line at π to approximate f (3), then yes, Δ x will be negative.
We'll wait to round until the very end of the problem. Now let's figure out the other values we need.
yold = f (π) = sin(5π) = 0.
Since
f '(x) = 5cos (5x)
the slope of the tangent line is
f '(π) = 5cos (5π) = -5.
Now we can do the estimation, leaving all rounding until the end:
The function y = f (x) is a solution to the IVP
Estimate f (0.5).
Answer
We have yold = 2, Δ x = 0.5, and
So
The function y = f (x) is a solution to the IVP
Estimate f (-½).
Answer
We have yold = 3, Δ x = -.5, and
So
The function y = f (x) is a solution to the IVP
Estimate f (2.01).
Answer
We have yold = 5, Δ x = .01, and
So
The function y = f (x) is a solution to the IVP
Approximate f (6.95).
Answer
We have yold = 3, Δ x = -.05, and
So
The function y = f (x) is a solution to the IVP
Approximate f (3.01).
Answer
We have yold = 5, Δ x = .01, and
So
