Find f if f''(x)=2+cos(x)
f(0)=5
f(pi/2)=−3
Well, let's see if I can come up with a funny response for you!
Why did the function go to therapy?
Because it couldn't handle the pressure of being f''(x)!
But in all seriousness, let's solve this equation. If we integrate f''(x), we get f'(x) = 2x + sin(x) + C1, where C1 is a constant. And if we integrate f'(x), we get f(x) = x^2 + cos(x) + C1x + C2, where C2 is another constant.
To find the values of C1 and C2, we can use the initial conditions. We know that f(0) = 5, so plugging that in, we get 0 + cos(0) + C1(0) + C2 = 5. Simplifying, we find that C2 = 5.
Next, we know that f(pi/2) = -3, so plugging that in, we get (pi/2)^2 + cos(pi/2) + C1(pi/2) + 5 = -3. Simplifying, we find that (pi/2)^2 + C1(pi/2) = -8.
From here, you would need to solve for C1 by plugging in the value of (pi/2) from the equation above. Once you find the value of C1, you can plug back into the equation for f(x) to get your final answer.
I hope that helps, and remember, math can be fun!
To find f(x), we can integrate the given equation twice.
First, integrate f''(x) with respect to x to obtain f'(x):
∫(f''(x)) dx = ∫(2 + cos(x)) dx
f'(x) = 2x + sin(x) + C1, where C1 is the constant of integration.
Next, integrate f'(x) with respect to x to find f(x):
∫(f'(x)) dx = ∫(2x + sin(x) + C1) dx
f(x) = x^2 + ∫sin(x) dx + C1x + C2
To evaluate the integral of sin(x), we can use the identity:
∫sin(x) dx = -cos(x) + C3, where C3 is the constant of integration.
Substituting this into the equation for f(x):
f(x) = x^2 - cos(x) + C3x + C2
Now, we can use the given initial conditions to find the specific values of the constants C2 and C3.
Using f(0) = 5:
5 = 0^2 - cos(0) + C3(0) + C2
5 = -1 + C2
So, C2 = 6.
Using f(pi/2) = -3:
-3 = (pi/2)^2 - cos(pi/2) + C3(pi/2) + 6
-3 = (pi^2)/4 + C3(pi/2) + 6
Simplifying:
(pi^2)/4 + C3(pi/2) = -9
We can solve this equation for C3:
C3(pi/2) = -9 - (pi^2)/4
C3(pi/2) = (-36 - pi^2)/4
C3 = (-36 - pi^2)/(4(pi/2))
C3 = (-36 - pi^2)/(2pi)
Substituting the values of C2 and C3 back into the equation for f(x):
f(x) = x^2 - cos(x) + (-36 - pi^2)/(2pi)x + 6
Therefore, f(x) = x^2 - cos(x) + (-36 - pi^2)/(2pi)x + 6.
To find f(x) given f''(x) = 2 + cos(x) and the initial conditions f(0) = 5 and f(pi/2) = -3, we need to integrate the equation twice.
Step 1: Integrate the equation f''(x) = 2 + cos(x) with respect to x to find f'(x):
∫f''(x) dx = ∫(2 + cos(x)) dx
f'(x) = ∫2 dx + ∫cos(x) dx
f'(x) = 2x + sin(x) + C₁,
where C₁ is the constant of integration.
Step 2: Integrate f'(x) to find f(x):
∫f'(x) dx = ∫(2x + sin(x) + C₁) dx
f(x) = ∫2x dx + ∫sin(x) dx + ∫C₁ dx
f(x) = x² + (-cos(x)) + C₁x + C₂,
where C₂ is the constant of integration.
Step 3: Use the initial conditions f(0) = 5 and f(pi/2) = -3 to solve for the constants C₁ and C₂.
For f(0) = 5:
5 = 0² + (-cos(0)) + C₁(0) + C₂
5 = -1 + C₂
C₂ = 6.
For f(pi/2) = -3:
-3 = (pi/2)² + (-cos(pi/2)) + C₁(pi/2) + 6
-3 = (pi/2)² + 0 + (pi/2)C₁ + 6
(pi/2)C₁ = -((pi/2)² + 9)
C₁ = -2(pi/2) - 18/(pi/2).
Step 4: Replace the values of C₁ and C₂ into the expression for f(x):
f(x) = x² + (-cos(x)) + C₁x + C₂
f(x) = x² - cos(x) + [-2(pi/2) - 18/(pi/2)]x + 6.
Therefore, f(x) = x² - cos(x) - 2(pi/2)x - 18/(pi/2)x + 6 is the solution.
f"(x) = 2+cos(x)
f'(x) = 2x + sin(x) + c1
f(x) = x^2 - cos(x) + c1*x + c2
f(0) = -1+c2 = 5 --> c1=6
f(π/2) = π^2/4 + c1*π/2 + 6 = -3
f(x) = x^2 - cos(x) + (π/2 - 18/π)x + 6