Suppose f′′(x)=−36sin(6x) and
f′(0)=−5, and f(0)=−3
f(π/3)=?
To evaluate f(π/3), we can use integration to find f(x) given the information about f'(x).
Step 1: Find f'(x)
Given f′′(x) = -36sin(6x), we can integrate f′′(x) to find f′(x).
∫f′′(x)dx = ∫(-36sin(6x))dx
Using the power rule of integration, we get:
f′(x) = 36/6cos(6x) + C1
f′(x) = 6cos(6x) + C1
Step 2: Find f(x)
Given f(0) = -3 and f′(x) = 6cos(6x) + C1, we can integrate f′(x) to find f(x).
∫f′(x)dx = ∫(6cos(6x) + C1)dx
Using the power rule of integration, we get:
f(x) = 6/6sin(6x) + C1x + C2
f(x) = sin(6x) + C1x + C2
Since f(0) = -3, we can substitute x = 0 and solve for C2.
sin(6(0)) + C1(0) + C2 = -3
C2 = -3
Therefore, the equation for f(x) becomes:
f(x) = sin(6x) + C1x - 3
Step 3: Evaluate f(π/3)
To evaluate f(π/3), substitute x = π/3 into the equation for f(x).
f(π/3) = sin(6(π/3)) + C1(π/3) - 3
simplifying gives:
f(π/3) = sin(2π) + C1(π/3) - 3
Since sin(2π) = 0, the equation simplifies further:
f(π/3) = 0 + C1(π/3) - 3
Now we need additional information about the value of C1 to find the exact value of f(π/3).
To find the value of f(π/3), we need to integrate the given function f′′(x) twice.
First, integrate f′′(x) to obtain f′(x):
∫ -36sin(6x) dx = -36 * (-1/6)cos(6x) + C₁ = 6cos(6x) + C₁
Using the condition f′(0) = -5, we can substitute x = 0 and solve for C₁:
6cos(6(0)) + C₁ = -5
6cos(0) + C₁ = -5
6(1) + C₁ = -5
6 + C₁ = -5
C₁ = -11
Therefore, f′(x) = 6cos(6x) - 11.
Second, integrate f′(x) to obtain f(x):
∫ (6cos(6x) - 11) dx = 6 * (1/6)sin(6x) - 11x + C₂ = sin(6x) - 11x + C₂
Using the condition f(0) = -3, we can substitute x = 0 and solve for C₂:
sin(6(0)) - 11(0) + C₂ = -3
sin(0) + C₂ = -3
0 + C₂ = -3
C₂ = -3
Therefore, f(x) = sin(6x) - 11x - 3.
Finally, we can find f(π/3) by substituting x = π/3 into the equation for f(x):
f(π/3) = sin(6(π/3)) - 11(π/3) - 3
= sin(2π) - (11π/3) - 3
= 0 - (11π/3) - 3
= -11π/3 - 3.
f" = -36sin(6x)
f' = 6cos(6x)+c
f'(0) = 6+c = -5, so c = -11
f' = 6cos(6x)-11
f = sin(6x)-11x+c
f(0) = c = -3
f = sin(6x)-11x-3
f(π/3) = sin(2π)-11(π/3)-3
= -11π/3 - 3