suppose f^n (x)= -36 sin(6x) and f'(0)=6 and f(0)= -5
f(π /4)= ?
To find f(π/4), we need to integrate f'(x).
Since f'(0) = 6, we can find f(x) by integrating this derivative:
f(x) = ∫6 dx = 6x + C
Using the information that f(0) = -5, we can solve for C:
-5 = 6(0) + C
C = -5
So, f(x) = 6x - 5.
Now, let's find f(π/4) using this equation:
f(π/4) = 6(π/4) - 5
f(π/4) = 3π/2 - 5
To find the value of f(π/4), we need to use the given information and apply the derivative property multiple times.
Given:
f^n(x) = -36 sin(6x)
f'(0) = 6
f(0) = -5
Let's start with the first derivative:
f'(x) = ((d/dx)^n)[f(x)]
Differentiating both sides of the equation f^n(x) = -36 sin(6x) with respect to x yields:
f^(n+1)(x) = -36 * (d/dx)[sin(6x)]
Since the derivative of sin(6x) with respect to x is 6cos(6x), the equation becomes:
f^(n+1)(x) = -36 * 6cos(6x)
Now, let's consider the given information:
f'(0) = 6
Substituting x = 0 into the equation for the first derivative:
f^(n+1)(0) = -36 * 6cos(6(0))
f^(n+1)(0) = -36 * 6cos(0)
f^(n+1)(0) = -36 * 6 * 1
f^(n+1)(0) = -216
Next, let's find the value of f(x) at x = 0:
f(0) = -5
We can write f(x) as:
f(x) = f(0) + ∫[0, x] f'(t) dt
Integrating f'(t) with respect to t gives:
∫[0, x] f'(t) dt = ∫[0, x] -36 * 6cos(6t) dt
Evaluating the integral gives:
f(x) = f(0) -36 * 6 * ∫[0, x] cos(6t) dt
f(x) = -5 -36 * 6 * ∫[0, x] cos(6t) dt
Now, we can find the value of f(π/4) by substituting x = π/4:
f(π/4) = -5 -36 * 6 * ∫[0, π/4] cos(6t) dt
To evaluate the integral, we need to use the antiderivative of cos(6t), which is sin(6t)/6:
f(π/4) = -5 -36 * 6 * [sin(6t)/6] evaluated from t = 0 to t = π/4
f(π/4) = -5 -36 * [sin(6π/4)/6 - sin(6(0))/6]
Simplifying further:
f(π/4) = -5 -36 * [sin(3π/2)/6 - sin(0)/6]
f(π/4) = -5 -36 * [-1/6]
Finally, evaluating the expression:
f(π/4) = -5 + 6 = 1
Therefore, f(π/4) = 1.