What is the local linearization of e^x^2 near x=1?
Is the answer 1+2x
To find the local linearization of a function near a point, we need to use the concept of the tangent line. The tangent line at a specific point represents the best linear approximation of the function near that point.
To find the local linearization of the function f(x) = e^(x^2) near x = 1, we can use the derivative of the function. The derivative of e^(x^2) can be found using the chain rule, which states that if we have a composition of functions, we can differentiate them by first differentiating the outer function and then multiplying it by the derivative of the inner function.
Using this rule, we can find the derivative of e^(x^2) as follows:
f'(x) = 2x * e^(x^2)
Next, we find the value of the function and its derivative at x = 1:
f(1) = e^(1^2) = e^1 = e
f'(1) = 2(1) * e^(1^2) = 2e
Now we have the point (1, e) on the function and its derivative at that point.
Finally, we can use the point-slope form of a line to express the local linearization:
y - y1 = m(x - x1)
Applying this formula with the values we obtained, we get:
y - e = 2e(x - 1)
Simplifying this equation, we can expand and rearrange to find the local linearization:
y = 2ex - 2e + e
Combining like terms, we get:
y = 2ex - e
Therefore, the local linearization of f(x) = e^(x^2) near x = 1 is given by y = 2ex - e.
To find the local linearization of a function, we can use the formula:
L(x) = f(a) + f'(a) * (x - a)
In this case, the function is f(x) = e^(x^2), and we want to find the local linearization near x = 1.
First, let's find the value of f(1) by substituting x = 1 into the function:
f(1) = e^(1^2) = e^1 = e
Next, let's find the derivative of f(x).
f'(x) = d/dx (e^(x^2))
Using the Chain Rule, the derivative of e^(x^2) with respect to x is 2x * e^(x^2).
So, f'(x) = 2x * e^(x^2).
Now, let's find the value of f'(1) by substituting x = 1 into the derivative:
f'(1) = 2(1) * e^(1^2) = 2e
Finally, we can use these values to find the local linearization L(x):
L(x) = f(1) + f'(1) * (x - 1)
= e + 2e * (x - 1)
= (1 + 2x)e - 2e
Therefore, the local linearization of e^(x^2) near x = 1 is (1 + 2x)e - 2e.
No, that is not the answer.
Use a McLaurin series about x = 1
f(x) = e at x = 1
f'(x) = 2x*e^x^2 which = 2e at x = 1
f(x) = f(1) + f'(1)(x-1) + ...
= e + 2e(x-1)