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
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.
1 answer
1 answer