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

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f(x) = (1+x)^-.5
f'(x) = -.5(1+x)^-1.5
series expansion(Taylor, Maclaurin)
f(x) = f(0) + f'(x)*x/1! + ......
f(x) = (1/2)(1+0)^-1.5 - .5(1+0)^-1.5 *x^1 + ...
= 1/2 - x/2 + ...
= (1-x)/2
Now try the others yourself.

To show that 1 - x/2 is the tangent line approximation to 1/√(1+x) near x=0, we need to find the equation of the tangent line and demonstrate that it is approximately equal to 1/√(1+x) for small values of x.

1. Find the derivative of the function f(x) = 1/√(1+x) using the chain rule. The derivative is given by:

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

2. Evaluate f'(0) to find the slope of the tangent line at x=0:

f'(0) = (-1/2)(1+0)^(-3/2) = -1/2

3. The equation of the tangent line at x=0 is given by:

y = f(0) + f'(0)(x - 0)

4. Substitute the value of f(0):

y = 1/√(1+0) - (1/2)(x)

Simplifying:

y = 1 - x/2

This demonstrates that 1 - x/2 is the tangent line approximation to 1/√(1+x) near x=0.

Now let's move on to the other questions:

2. To find the local linearization of e^(x^2) near x=1, we need to determine a linear function that approximates the behavior of the function e^(x^2) near x=1.

1. Start by finding the first derivative of the function f(x) = e^(x^2):

f'(x) = 2xe^(x^2)

2. Evaluate f'(1) to find the slope of the tangent line at x=1:

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

3. The equation of the tangent line at x=1 is given by:

y = f(1) + f'(1)(x - 1)

4. Substitute the value of f(1):

y = e^(1^2) + 2e(x - 1)

Simplifying:

y = e + 2e(x - 1)

This equation represents the local linearization of e^(x^2) near x=1.

3. To find the tangent line approximation of 1/x near x=1:

1. Consider the function f(x) = 1/x.

2. Find the derivative of f(x) using the power rule:

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

3. Evaluate f'(1) to find the slope of the tangent line at x=1:

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

4. The equation of the tangent line at x=1 is given by:

y = f(1) + f'(1)(x - 1)

5. Substitute the value of f(1):

y = 1/1 + (-1)(x - 1)

Simplifying:

y = 1 - (x - 1)

y = 2 - x

This equation represents the tangent line approximation of 1/x near x=1.