Math, asked by prathammali249, 3 months ago

1.1 x0.1 x .01 = please find​

Answers

Answered by durgeshjatode1
0

Step-by-step explanation:

Consider a function f that is differentiable at a point x=a. Recall that the tangent line to the graph of f at a is given by the equation

y=f(a)+f^{\prime}(a)(x-a).

For example, consider the function f(x)=\frac{1}{x} at a=2. Since f is differentiable at x=2 and f^{\prime}(x)=-\frac{1}{x^2}, we see that f^{\prime}(2)=-\frac{1}{4}. Therefore, the tangent line to the graph of f at a=2 is given by the equation

y=\frac{1}{2}-\frac{1}{4}(x-2).

(Figure)(a) shows a graph of f(x)=\frac{1}{x} along with the tangent line to f at x=2. Note that for x near 2, the graph of the tangent line is close to the graph of f. As a result, we can use the equation of the tangent line to approximate f(x) for x near 2. For example, if x=2.1, the y value of the corresponding point on the tangent line is

y=\frac{1}{2}-\frac{1}{4}(2.1-2)=0.475.

The actual value of f(2.1) is given by

f(2.1)=\frac{1}{2.1}\approx 0.47619.

Therefore, the tangent line gives us a fairly good approximation of f(2.1) ((Figure)(b)). However, note that for values of x far from 2, the equation of the tangent line does not give us a good approximation. For example, if x=10, the y-value of the corresponding point on the tangent line is

y=\frac{1}{2}-\frac{1}{4}(10-2)=\frac{1}{2}-2=-1.5,

whereas the value of the function at x=10 is f(10)=0.1.

This figure has two parts a and b. In figure a, the line f(x) = 1/x is shown with its tangent line at x = 2. In figure b, the area near the tangent point is blown up to show how good of an approximation the tangent is near x = 2.

Figure 1. (a) The tangent line to f(x)=1/x at x=2 provides a good approximation to f for x near 2. (b) At x=2.1, the value of y on the tangent line to f(x)=1/x is 0.475. The actual value of f(2.1) is 1/2.1, which is approximately 0.47619.

In general, for a differentiable function f, the equation of the tangent line to f at x=a can be used to approximate f(x) for x near a. Therefore, we can write

f(x)\approx f(a)+f^{\prime}(a)(x-a) for near [latex]a.

We call the linear function

L(x)=f(a)+f^{\prime}(a)(x-a)

the linear approximation, or tangent line approximation, of f at x=a. This function L is also known as the linearization of f at x=a.

To show how useful the linear approximation can be, we look at how to find the linear approximation for f(x)=\sqrt{x} at x=9.

Answered by Anonymous
3

Step-by-step explanation:

1.1 x 0.1 x .01 = 0.0011

Hope it helps.

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