Using Tangent Lines to Approximate Function Values Examples

Estimate sin(3) using a tangent line approximation at π.

3 is close to π ≈ 3.14, so it makes sense to find the tangent line to f(x) = sin(x) at π and use that to approximate sin(3):

We'll give a freebie right now by revealing that f '(π) = -1. We know f (π) = sin(π) = 0. Now we can build the tangent line equation:

y = mx + b

y = (-1)x + b

0 = (-1)(π) + b

b = π

so

y = (-1)x + π

Now we find the y-value on the tangent line when x = 3:

(-1)(3) + π ≈ 0.14.

This is close to sin(3).

Yes, it's possible to plug "sin(3)" into a calculator and find an answer. A lot of people don't realize the calculator is doing essentially the same thing we are: it's estimating. It's doing something more like a tangent line approximation on a caffeine high, and it's doing it faster than we can, but it's still estimating.

If the end of the world comes and all the calculators are destroyed, we'll still be able to estimate sin(3). We hope that's a reassuring thought.

Use a tangent line approximation to estimate (0.9)³.

We're looking at the function f(x) = x³, and we want the tangent line near x = 1. We found earlier that the equation of this tangent line is

y = 3x – 2.

To estimate, we plug in 0.9 for x in the tangent line equation and see what we find:

3(0.9) – 2 = 2.7 – 2 = 0.7.

Our estimation is 0.7. The calculator says (0.9)³ = 0.729, so we're close.

There are a bunch of phrases that all mean the same thing. If we're asked to use a local linearization or a linear approximation to estimate a function value, these mean "do a tangent line approximation." They're all different ways of saying "use a line to approximate the function."

Use a local linearization to estimate f(0.8) if .

Since f(1) = 1 doesn't require much thinking, and 0.8 is close to 1, we'll use the tangent line to f at 1 for this approximation. We found earlier that the equation of the tangent line is

y = -2x + 3.

Plugging in 0.8 for x, we see that

f(0.8) ≈ -2(0.8) + 3 = 1.3.

The actual value of f(0.8) is 1.5625 so we're a little off, but still close.

Use a tangent line approximation to estimate f(-1.2) where f(x) = 1 – x².

We know f(-1) = 0, so that's a good place to draw the tangent line. We found earlier that the equation of the tangent line is

y = 2x + 2.

We'll put in -1.2 for x to find

f(-1.2) ≈ 2(-1.2) + 2 = -0.4.

Use a linear approximation to estimate f(0.035) for f(x) = 2x³.

0.035 is close to 0, and f(0) = 0 is possible to find, so now let's find the tangent line at 0. We need to calculate f ' (0) first.

The equation for the tangent line is

y = 0 + 0(x – 0) = 0.

That's right, the equation for the tangent line is the boring horizontal line y = 0. Therefore

f(0.035) ≈ 0.

Since 0.035 is small and only gets smaller when cubed, this estimate is a good one.

We'll find more interesting approximation questions after we talk about how to find derivatives of more complicated functions.