Consider the function f(x) whose second derivative is f''(x)=5x+2sin(x). If f(0)=3 and f'(0)=3, what is f(3)?
f" = 5x+2sinx
f' = 5/2 x^2 - 2cosx + c
f'(0)=3, so
0-2+c=3
c=5
so, f' = 5/2 x^2 - 2cosx + 5
f = 5/6 x^3 - 2sinx + 5x + c
f(0)=3, so
0-0+0+c=3
c=3
so, f=5/6 x^3 - 2sinx + 5x + 3
I assume you can now find f(3)
To find the value of f(3), we can start by integrating the given second derivative, f''(x), to find the first derivative f'(x):
∫(f''(x)) dx = ∫(5x + 2sin(x)) dx
Using the power rule of integration and the integral of sin(x), we can solve this as follows:
f'(x) = (5/2)x^2 - 2cos(x) + C1
Next, we can use the initial condition f'(0) = 3 to find the value of the constant C1:
f'(0) = (5/2)(0)^2 - 2cos(0) + C1
3 = 0 - 2(1) + C1
3 = -2 + C1
C1 = 5
Now that we have the first derivative f'(x), we can integrate it to find the original function f(x):
∫(f'(x)) dx = ∫((5/2)x^2 - 2cos(x) + C1) dx
Using the power rule of integration and the integral of cos(x), we can solve this as follows:
f(x) = (5/6)x^3 - 2sin(x) + C1x + C2
Next, we can use the initial condition f(0) = 3 to find the value of the constant C2:
f(0) = (5/6)(0)^3 - 2sin(0) + C1(0) + C2
3 = 0 - 0 + 0 + C2
C2 = 3
Now we have the equation for f(x):
f(x) = (5/6)x^3 - 2sin(x) + 5x + 3
To find f(3), we substitute x = 3 into the equation:
f(3) = (5/6)(3)^3 - 2sin(3) + 5(3) + 3
Simplifying this expression will give us the value of f(3).