find the linearization L(x) of the function a. f(x)=sin(x), a=pi,3
Consider the function used to find the linearization at a.
L(x) = f(a) + f'(a) (x-a)
Substitute the value of a=π/3 into the linearization function.
L(x) = f(π/3) + f'(π/3) (x-π/3)
Evaluate f(π/3) which is √3/2
Find the derivative and evaluate it at a = π/3 which is 1/2
Substitute the components into the linearization function in order to find the linearization at a.
L(x) = √3/2 + 1/2 (x- √3/2)
Simplify each term.
L(x) = √3/2 + x/2 - π/6
Why did the function go to the circus? Because it wanted to find its linearization and join the line-up of funny functions. Now, let's get to business.
The linearization of a function f(x) at a point a is given by the equation L(x) = f(a) + f'(a)(x - a), where f'(a) represents the derivative of f(x) at x = a.
For f(x) = sin(x) and a = π, we need to find f'(a) first. The derivative of sin(x) is cos(x), so f'(x) = cos(x).
Now, let's find f'(a) = cos(π):
f'(π) = cos(π) = -1.
Thus, the linearization L(x) of f(x) = sin(x) at a = π is:
L(x) = f(π) + f'(π)(x - π)
= sin(π) + (-1)(x - π)
= 0 - (x - π)
= π - x.
So, the linearization L(x) of f(x) = sin(x) at a = π is L(x) = π - x. Ta-da!
To find the linearization L(x) of the function f(x) = sin(x) at a given point a, we need to use the point-slope form of a linear equation. The formula for the linearization at a is:
L(x) = f(a) + f'(a)(x - a)
where f(a) represents the value of the function at a, and f'(a) represents the derivative of the function evaluated at a.
In this case, a is given as pi and 3. Let's find the linearization for each case:
Case 1: a = pi
To find the linearization at a = pi, we first need to evaluate f(pi) and f'(pi).
f(pi) = sin(pi) = 0
f'(pi) = cos(pi) = -1
Now we can substitute these values into the linearization formula:
L(x) = f(pi) + f'(pi)(x - pi)
= 0 + (-1)(x - pi)
= -x + pi
Therefore, the linearization of f(x) = sin(x) at a = pi is L(x) = -x + pi.
Case 2: a = 3
To find the linearization at a = 3, we need to evaluate f(3) and f'(3).
f(3) = sin(3) ≈ 0.141
f'(3) = cos(3) ≈ -0.989
Substituting these values, we get:
L(x) = f(3) + f'(3)(x - 3)
≈ 0.141 + (-0.989)(x - 3)
≈ -0.989x + 2.917
Therefore, the linearization of f(x) = sin(x) at a = 3 is L(x) = -0.989x + 2.917.
impatient much?
still can't type fractions?