Suppose f”(x) =-36sin(6x) and f’(0)=-5 and f(0)=-6 find f(pi/5)
f" = -36sin(6x)
f' = 6cos(6x)+c
f'(0) = 6+c = -5 --> c = -11
f' = 6cos(6x)-11
f = sin(6x) - 11x + c
f(0) = 0 - 0 + c = -6
f = sin(6x)-11x-6
f(π/5) = sin(6π/5) - 11π/5 - 6 = -13.5
To find f(pi/5) given f''(x) = -36sin(6x), f'(0) = -5, and f(0) = -6, you can use integration to find f(x) given f''(x).
First, integrate the second derivative -36sin(6x) twice to find the expression for f(x). We will integrate twice since the second derivative implies f(x) is the integral of the first derivative, and the first derivative is the integral of the original function, f''(x).
When integrating -36sin(6x) twice, we bring back the constant term -36, and integrate sin(6x) twice.
∫(-36sin(6x)) dx = -36 ∫sin(6x) dx
The integral of sin(6x) can be found by applying the chain rule. Let u = 6x, then du = 6 dx or dx = du/6.
∫sin(6x) dx = ∫sin(u) (du/6) = (1/6) ∫sin(u) du
Now, integrate sin(u) to get:
(1/6) ∫sin(u) du = (-1/6)cos(u) + C
Substituting back u = 6x:
(-1/6)cos(6x) + C₁
Now, integrate again:
∫((-1/6)cos(6x))dx = (-1/6) ∫cos(6x)dx
Using the same u-substitution as before, where u = 6x:
(-1/6) ∫cos(6x)dx = (-1/6) ∫cos(u) (du/6) = (-1/36) ∫cos(u) du
Integrating cos(u) gives:
(-1/36) ∫cos(u) du = (-1/36)sin(u) + C₂
Substituting back u = 6x:
(-1/36)sin(6x) + C₂
Finally, we have the expression for f(x):
f(x) = (-1/36)sin(6x) + C₂
To determine the value of C₂, we can substitute the initial conditions given:
f'(0) = -5, which means the derivative of f(x) when x = 0 is -5:
f'(0) = (-1/36)sin(6(0)) + C₂
-5 = (-1/36)(0) + C₂
C₂ = -5
So the expression for f(x) is now:
f(x) = (-1/36)sin(6x) - 5
Next, we can find f(x) when x = pi/5:
f(pi/5) = (-1/36)sin(6(pi/5)) - 5
Simplifying further:
f(pi/5) = (-1/36)sin(6/5 * pi) - 5
Since sin(6/5 * pi) = sin(pi + pi/5) = -sin(pi/5), we can write:
f(pi/5) = (-1/36)(-sin(pi/5)) - 5
f(pi/5) = (1/36)sin(pi/5) - 5
Hence, f(pi/5) = (1/36)sin(pi/5) - 5.