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.