Find the linearization of the function below at x = π/3.?
y=sin(x)/x
Your answer should be a linear function of x where the coefficients are accurate to at least two decimal places. You can enter π in your answer as "pi" (without the quotes).
I am not able to figure out how to find the answer. I tried couple times but it is showing incorrect.Please show the answer as well because I was not able to get the correct answer last time either. Please help me.. Thank you.
take y' = dy / dx
y' = [x cos x -sin x]/x^2
at x = pi/3
y' = [ pi/3 cos pi/3 - sin pi/3 ] / (pi^2/9)
=[ pi/3 *1/2 - sqrt3 /2 ]/(pi^2/9)
= [ pi/6 - 3 sqrt 3 /6 ] 9/pi^2
= 1.5(pi-3 sqrt 3)/pi^2
that is m in y = mx+b
to get b once you calculate m
at x = pi/3
y = sin (pi/3) / (pi/3)
y = (3/pi) sqrt3 / 2
so
1.5 sqrt3/ pi = m (pi/3) + b
solve for b
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(x) is the original function, a is the point at which the linearization is being evaluated, and f'(x) is the derivative of the original function.
In this case, the given function is y = sin(x)/x. To find the linearization at x = π/3, we need to calculate f(π/3) and f'(π/3).
1. First, let's calculate f(π/3):
Substitute x = π/3 in the given function:
f(π/3) = sin(π/3)/(π/3) = (√3/2)/(π/3) = (√3π)/(2π/3) = 3√3/2
2. Next, let's calculate f'(π/3):
Take the derivative of the given function:
f(x) = sin(x)/x
To find the derivative, we can use the quotient rule:
f'(x) = (x*cos(x) - sin(x))/x^2
Now, substitute x = π/3 in the derivative:
f'(π/3) = (π/3)*cos(π/3) - sin(π/3))/(π/3)^2
= (π/3)*(1/2) - (√3/2))/(π^2/9)
= (π/6 - √3/2)/(π^2/9)
= (9π/6 - 3√3)/(π^2/9)
= 3π/2π^2 - 3√3/(π^2/9)
= 3/2π - 3√3/π^2 * 9/1
= 27/2π - 27√3/π^2
3. Now, let's substitute the values into the linearization formula:
L(x) = f(π/3) + f'(π/3)(x - π/3)
L(x) = 3√3/2 + (27/2π - 27√3/π^2)(x - π/3)
L(x) = 3√3/2 + (27/2π - 27√3/π^2)(x - π/3)
L(x) = 3√3/2 + (27/2π - 27√3/π^2)x - (27/2π - 27√3/π^2)π/3
Therefore, the linearization of the function y = sin(x)/x at x = π/3 is:
L(x) = 3√3/2 + (27/2π - 27√3/π^2)x - (27/2π - 27√3/π^2)π/3
Now, you can simplify this expression further if needed or use it to calculate the linearized value at any given x.