Find the linearization of f(x)= sqrtx at x=25 and use the linearization to approximate sqrt25.2
This is my math: 5+ 1/10 * (25.2-25) but it's wrong.
The McLaurin series expansion of that function about x=25 is
f(x) = f(25) + (x-25)[f'(x) at x=25)
f'(x) = (1/2)*x^(-1/2)
f'(25) = (1/2)/5 = 1/10
Therefore
f(x) = 5 + (x-25)/10
f(25.2) = 5+ .02 = 5.02
There is nothing wrong with your answer. The correct value is
5.01996016...
Maybe they wanted you to do the last step and compute an actual numerical approximation, 5.02.
I typed in 5.01999999999 and it took it. Weird.
To find the linearization of a function at a given point, we need to use the formula:
L(x) = f(a) + f'(a)(x - a)
Where L(x) is the linearization function, f(a) is the value of the function at the point a, f'(a) is the derivative of the function evaluated at the point a, and (x - a) represents the distance between the point x and a.
In this case, we are given the function f(x) = sqrt(x), and we want to find the linearization at x = 25.
Step 1: Find the value of f(25)
To find f(25), simply plug in x = 25 into the function:
f(25) = sqrt(25) = 5
Step 2: Find the derivative f'(x)
To find the derivative of the function f(x) = sqrt(x), we can use the power rule for differentiation:
f'(x) = (1/2) * x^(-1/2) = 1/(2sqrt(x))
Step 3: Find the derivative f'(25)
To find f'(25), simply plug in x = 25 into the derivative function:
f'(25) = 1/(2sqrt(25)) = 1/10
Step 4: Plug the values into the linearization formula
Now we have all the necessary values to find the linearization L(x):
L(x) = f(a) + f'(a)(x - a)
L(x) = 5 + (1/10)(x - 25)
To approximate the value of sqrt(25.2) using the linearization, substitute x = 25.2 into the linearization function:
L(25.2) = 5 + (1/10)(25.2 - 25)
L(25.2) = 5 + (1/10)(0.2)
L(25.2) = 5 + 0.02
L(25.2) = 5.02
Therefore, the linearization of f(x) = sqrt(x) at x = 25 is L(x) = 5 + (1/10)(x - 25), and the approximation for sqrt(25.2) using the linearization is 5.02.