Find f.
f ''(x) = 6 + cos x, f(0) = −1,
f(7π/2) = 0
Find f.
f ''(x) = 6 + cos x, f(0) = −1, f(9π/2) = 0
Well, let's have some fun finding f! To find f, we'll start by integrating f ''(x) to get f'(x). The integral of 6 is 6x, and the integral of cos x is sin x. So f'(x) = 6x + sin x.
Now, we need to find the constant of integration. We can do that by using the given information f(0) = -1. Plugging in x = 0 into f'(x) gives us 0 + sin(0) = 0. So the constant of integration is -1.
Now let's find f(x) by integrating f'(x). The integral of 6x is 3x^2, and the integral of sin x is -cos x. Adding the constant of integration -1, we get f(x) = 3x^2 - cos x - 1.
But we're not quite done yet! We also have the information f(7π/2) = 0. Plugging in x = 7π/2 into our equation for f(x), we get 0 = 3(7π/2)^2 - cos(7π/2) - 1. Well, I'm not going to evaluate that mess, but you can do it if you want! Just rearrange the equation to solve for π and calculate away.
So voila! We found f(x) as 3x^2 - cos x - 1. Just remember to solve for π to find the value of f(7π/2). Good luck, mathlete!
To find the function f, we will integrate the given second derivative of f.
1. First, integrate the given second derivative: f'(x) = ∫ (6 + cos x) dx.
The integral of 6 with respect to x is 6x, and the integral of cos x with respect to x is sin x. Therefore, f'(x) = 6x + sin x + C1, where C1 is a constant.
2. Next, integrate the obtained first derivative to find f(x): f(x) = ∫ (6x + sin x + C1) dx.
The integral of 6x with respect to x is 3x^2, and the integral of sin x with respect to x is -cos x. Integrating C1 gives C1x. Therefore, f(x) = 3x^2 - cos x + C1x + C2, where C2 is a constant.
3. Use the given initial conditions to determine the values of the constants C1 and C2.
Given f(0) = -1, we can substitute x = 0 in the equation: -1 = 3(0)^2 - cos(0) + C1(0) + C2.
This simplifies to -1 = -1 + C2. This implies C2 = 0.
4. Next, use the second initial condition f(7π/2) = 0 to find the value of C1.
Substitute x = 7π/2 in the equation: 0 = 3(7π/2)^2 - cos(7π/2) + C1(7π/2) + 0.
Simplify and solve for C1.
Note that cos(7π/2) = 0, so the equation becomes 0 = 3(7π/2)^2 + C1(7π/2).
This gives us: 0 = 147π^2/2 + C1(7π/2).
Rearranging, we get: C1(7π/2) = -147π^2/2.
Divide both sides by 7π/2 to solve for C1: C1 = -147π/14.
5. Now, substitute the values of C1 and C2 back into the equation for f(x):
f(x) = 3x^2 - cos x - (147π/14)x.
So, the function f(x) is given by f(x) = 3x^2 - cos x - (147π/14)x.
f ''(x) = 6+cosx
f '(x) = 6x + sinx + c
f (x) = 3x^2 - cosx + cx + k
f(0) = 1
0 - cos 0 + 0 + k = 1
-1 +k= 1
k = 2
so f(x) = 3x^2 - cosx + cx + 2
f(7π/2) = 0
3(7π/2)^2 - cos(7π/2) + (7π/2)c +2 = 0
solve for c, plug into the f(x) equation and you got it