### Topics

## Introduction to Derivatives - At A Glance:

"Approximation" is what we do when we can't or don't want to find an exact value. We're going to approximate actual function values using tangent lines.

We pointed out earlier that if we zoom in far enough on a continuous function, it looks like a line. For example, take the function *f*(*x*) = *x*^{2} and zoom in around *x* = 1. If we zoom in enough near *x* = 1, the function *f* looks like a line.

If we graph the function and its tangent line at 1, we'll see that as we zoom in around *x* = 1, the function *f* looks like its tangent line.

If we zoom back out a little bit, the function doesn't look quite so much like a line. However, the function and its tangent line are still "close together."

This means, for example, that the *y*-value on the tangent line at x = 1.1 is "close" to the *y*-value on the function *f*(*x*) = *x*^{2} when *x* = 1.1.

We found earlier that the tangent line to *f*(*x*) = *x*^{2} at 1 has the equation:

y = 2*x*-1.

If we don't feel like calculating the actual value *f*(1.1), we can instead plug 1.1 into the tangent line equation and see what comes out:

2(1.1) - 1 = 2.2 - 1 = 1.2.

This is a good approximation to *f*(1.1):

If we then go and calculate the exact value of the function, we find

*f*(1.1) = 1.21.

This means our approximation was only 0.01 off.

Why bother? Approximation is supposed to make life easier, so why should we go to all that work of finding the equation of a line and finding the *y*-value of the line when x = 1.1 instead of calculating *f*(1.1) and being done with it?

In that example, we could calculate *f*(1.1) exactly. But we can't do that for every function.

#### Example 1

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 *3* and use that to approximate sin(3): We'll give a freebie right now that *f*'(π) = -1. We know *f*(π) = sin(π) = 0. Now we build the tangent line equation: *y* = *mx* + *b*
*y* = (-1)*x* + *b*
0 = (-1)(π) + *b* *b* = π
*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 calculate the sin of 3. We hope that's a reassuring thought. | |

#### Example 2

Use a tangent line approximation to estimate (0.9)^{3}.
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We're looking at the function *f*(*x*) = *x*^{3}, and we want the tangent line near *x* = 1. We found earlier that the equation of this tangent line is *y* = 3*x* - 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." | |

#### Example 3

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* = -2*x* + 3.
Plugging in 0.8 for *x*, we see that *f*(*x*) ≅ -2(0.8) + 3 = 1.4.
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#### Example 4

Use a tangent line approximation to estimate *f*(-1.2) where *f*(*x*) = 1 - *x*^{2}.
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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* = 2*x* + 2.
We put in -1.2 for *x* and find *f*(-1.2) ≅ 2(-1.2) + 2 = -0.4.
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#### Example 5

Use a linear approximation to estimate *f*(0.035) for *f*(*x*) = 2*x*^{3}.
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0.035 is close to 0, and *f*(0) = 0 is possible to find, 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. | |