Find the linear approximation at x = 0 to show that the...

Find the linear approximation at x = 0 to show that the following commonly used approximations are valid for "small" x. Compare the approximate and exact values for x = 0.01, x = 0.1, and x = 1. Round your calculations to seven decimal places if needed. tan(x) ? x L(x) f(x) x = 0.01 0.01 0 x = 0.1 0.1 0 x = 1 1 0 Note: f(x) = tan(x)

Transcript

00:01 So define our linear approximations, so we can compare them to the actual values of tangent.

00:07 We first need to fill everything out in this equation here, where a is the center, and in this case, it tells a is zero.

00:15 So i'm just going to go ahead and plug those in over here really quickly.

00:20 And that's going to simplify this down to f of 0 plus f prime of 0, and then that would just be times x.

00:28 So let's come over here and first find f of 0.

00:32 Well, f is tangent.

00:35 So f of 0 is going to be tangent of 0.

00:39 And tangent is sine over cosine, which would be 0 over 1, so that's just going to be 0.

00:44 Now we need to find the derivative.

00:48 And so the derivative of tangent should be secant squared.

00:57 And sikin is just 1 over cosine.

01:03 So plug it in 0 into here, so f prime of 0 is going to just be 1, because it would be 1 over 1 squared and that would just be 1.

01:16 So now we can come over here and plug all this in.

01:18 So l of x, our linear approximation, close to 0 is going to be, well, f of 0 is 0, f prime of 0 is 1, which is just going to leave us with x.

01:34 So now let's go ahead and plug in those values.

01:37 So we had, let's see, so for the x values, i believe it was 0 .01, 0 .1, and then just 1.

01:51 And then l of x, well, l of x, we already said it's just x.

01:56 So plugging those in would be 0 .01, 0 .1, and 1...

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