Use an appropriate function and local linear approximation to find an approximation of the given

step 1 Approximate 7 .9 to the two -thirds. And I'm not sure if the function is F of X is X to the two

Use an appropriate function and local linear approximation...

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Key Concepts

Derivative

The derivative of a function represents the rate at which the function's value changes with respect to changes in its input. In the context of local linear approximation, the derivative at a specific point provides the slope of the tangent line, which is crucial for constructing the linear model that approximates the function.

Differentiation of Power Functions

When dealing with power functions, such as functions in the form f(x) = x^(n), the power rule of differentiation is applied. This rule states that the derivative of x^(n) is n*x^(n-1). This process is essential for finding the slope of the tangent line used in the local linear approximation for functions involving powers.

Local Linear Approximation

Local linear approximation, or linearization, is a technique to estimate the value of a function near a known point using the tangent line at that point. It simplifies complex functions by approximating them with a linear function over a small interval, allowing for easier calculations and estimations.

Tangent Line

The tangent line at a point on a function is the straight line that best approximates the function near that point. Its slope is determined by the derivative of the function at the given point, and it is used as the basis for the local linear approximation.

Transcript

00:01 Approximate 7 .9 to the two -thirds.

00:03 And i'm not sure if the function is f of x is x to the two -thirds, or f of x equals 8 minus x to the two -thirds.

00:15 See, i'm going to do them both at the same time.

00:18 Okay, f of x is approximately equal to f of a plus f prime at a, x minus a.

00:29 So in the first one, a is eight, and the second one, a is 0.

00:37 So f of a is 8 to the 2 thirds, which is the cube root of 8 squared, 4.

00:44 And over here, f of a is 0 to the 2 thirds, which is 0.

00:52 F prime of x is 2 thirds, x to the minus 1 third, or 2 over 3, x to the 1 3.

01:02 And here, f prime of x is two thirds, eight minus x to the minus one third times minus one.

01:15 So minus two over three, eight minus x to the one third.

01:24 Okay, f prime at a, that's f prime at eight.

01:30 That is two over three times eight to the one third.

01:35 Cube root of 8 is 2 so that's 2 over 6 which is 1 3rd and here f is 0 i mean a is 0 so f prime at a is f prime at 0 which is minus 2 over 3 times 8 to the 1 3 which is minus 1 3 which is minus 1 3.

02:01 Okay, so f of x is approximately 4 plus one -third x minus 8.

02:14 And here, f of x is approximately 0 minus 1 -third x minus 0.

02:28 So minus 1 -third x for this one.

02:32 So then what were we trying to find? i forgot 7 .9...

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