Estimate √(16.2) using the linearization L(x) of f(x)=√(x) at a=16. Plot f and L on the same set

Estimate √(16.2) using the linearization L(x) of f(x)=√(x)...

Key Concepts

Graphical Analysis

Graphical analysis involves plotting both the original function and its linear approximation to visually compare their behaviors. This helps to understand the accuracy of the estimation and whether the approximation overestimates or underestimates the actual values.

Concavity

Concavity describes the curvature of a function. A function that is concave down has a curvature such that its tangent line lies above the graph of the function, indicating that linear approximations may overestimate the function's values.

Linearization

Linearization is the process of approximating a function near a specific point using the tangent line at that point. This concept uses the value of the function and its derivative at the point of tangency to create a simple linear function that approximates the original function for values close to that point.

Derivative

The derivative of a function at a particular point gives the slope of the tangent line at that point. It measures the rate at which the function's value changes and is crucial for constructing accurate linear approximations.

Tangent Line

The tangent line to a curve at a point is the straight line that touches the curve without crossing it in a small neighborhood around that point. It provides the best linear approximation of the function near the point of tangency.

Transcript

00:01 Okay, find the linear approximation for f of x equals the square root of x around a equals 16 and estimate the square root of 16 .2.

00:11 And then plot f and l and decide if l is bigger or smaller than the actual value.

00:18 Okay, so here's what i'm going to do.

00:22 I'm going to, i can't remember the formula exactly.

00:25 So i'm going to say the derivative of x is almost equal to the, slope of the tangent of the derivative sorry the derivative at a f of x minus f of a over x minus a okay now i'm going to cross multiple put this over one i'm going to cross multiply f prime of a times x minus a is approximately equal to f of x minus f of a and so f of x is approximately equal to i'm going to add f of a to both sides.

01:10 I just moved it to the other side.

01:12 F of a plus f prime of a x minus a.

01:17 That's the linearization formula.

01:20 Oops, l of x.

01:27 Okay, so f of x in this problem is the square root of x and a is 16.

01:39 So f of a is 16, which is the square root of 16, which is 4.

01:51 F prime of x, square root of x is x to the 1 half.

01:55 So f prime of x is 1 half x to the minus 1 half, which is 1 over 2 square roots of x.

02:04 So f prime of 16 is 1 over 2 square roots of 16, which is 1 over 2 times 4, which is 1 8th.

02:16 So l of x equals f of a, which is 4, plus the derivative which is one eighth times x minus a which is 16 so that's the linearization you can use it to approximate the square root of anything that's not too far away from 16 so let's use it to approximate the square root of 16 .2 so that's 4 plus 1 8th 16 .2 minus 16 that's 4 plus 1 8th of 0 .2 which i'm going to write as two tenths.

03:02 So that's four plus one eighth times one -fifth.

03:08 Four and one -fortieth.

03:13 Okay, so let's see, one -fourteenth.

03:16 If i multiply that by two and a half, that it gave me 100.

03:27 I'm just going to multiply it by 25 here.

03:29 That makes more sense...

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