1-show that 1-x/2 is the tangent line approximation to 1/ sqrt of 1+x near x=0

2-what is the local linearization of e^x^2 near x=1

3-find the tnagent line approximation of 1/x near x=1

1. To show that 1 - x/2 is the tangent line approximation to 1/sqrt(1+x) near x = 0, we can use the concept of Taylor series expansion. The Taylor series expansion of a function f(x) about a point a can be expressed as follows:

f(x) = f(a) + f'(a)*(x-a) + f''(a)*(x-a)^2/2! + f'''(a)*(x-a)^3/3! + ...

In this case, we want to approximate the function 1/sqrt(1+x) near x = 0.

First, let's find the derivative of the function:

f'(x) = -1/(2 * (1+x)^(3/2))

Now, let's evaluate the function and its derivative at x = 0:

f(0) = 1
f'(0) = -1/2

Substituting these values into the Taylor series expansion formula:

1/sqrt(1+x) ≈ 1 + (-1/2)(x-0)

Simplifying the expression:

1 - x/2

Therefore, we have shown that 1 - x/2 is the tangent line approximation to 1/sqrt(1+x) near x = 0.

2. To find the local linearization of e^x^2 near x = 1, we can once again use the concept of Taylor series expansion.

The Taylor series expansion of the function e^x about a point a can be expressed as follows:

e^x = e^a + e^a*(x-a) + e^a*(x-a)^2/2! + e^a*(x-a)^3/3! + ...

In this case, we want to approximate the function e^x^2 near x = 1.

First, let's evaluate the function and its derivative at x = 1:

f(1) = e^1 = e
f'(1) = 2x*e^x^2 = 2e^1 = 2e

Substituting these values into the Taylor series expansion formula:

e^x^2 ≈ e + 2e*(x-1)

Simplifying the expression:

e(1 + 2(x-1))

Therefore, the local linearization of e^x^2 near x = 1 is given by e(1 + 2(x-1)).

3. To find the tangent line approximation of 1/x near x = 1, we can use the concept of linear approximation.

The equation of a line can be represented by the formula y = mx + b, where m is the slope and b is the y-intercept.

First, let's find the derivative of the function f(x) = 1/x:

f'(x) = -1/x^2

Now, let's evaluate the derivative at x = 1:

f'(1) = -1/1^2 = -1

The slope of the tangent line is -1.

Next, let's evaluate the function at x = 1:

f(1) = 1/1 = 1

The point (1,1) lies on the tangent line.

Using the formula for a line, y = mx + b, we can substitute the slope and the point to solve for b:

1 = -1(1) + b
1 = -1 + b
b = 2

Therefore, the equation of the tangent line approximation of 1/x near x = 1 is y = -x + 2.