"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) = x2 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) = x2 when x = 1.1.
We found earlier that the tangent line to f(x) = x2 at 1 has the equation:
y = 2x – 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 of 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. Try doing this with a function like ex or ln(x). Without a calculator, evaluating those functions for most values of x will get pretty hairy.
Example 1
Estimate sin(3) using a tangent line approximation at π. |
Example 2
Use a tangent line approximation to estimate (0.9)3. |
Example 3
Use a local linearization to estimate f(0.8) if |
.Example 4
Use a tangent line approximation to estimate f(-1.2) where f(x) = 1 – x2. |
Example 5
Use a linear approximation to estimate f(0.035) for f(x) = 2x3. |
