Section Antiderivatives:
Find the function f(x) satisfying the given conditions.
f''(t) = 4 + 6t, f(1) = 3, f(-1) = -2
To find the function f(x) satisfying the given conditions, we will perform a two-step process. First, we'll find the antiderivative of f''(t) to find f'(t). Then, we'll find the antiderivative of f'(t) to obtain f(t). We will then use the initial conditions f(1) = 3 and f(-1) = -2 to determine the specific function f(x).
Step 1: Find f'(t):
To find f'(t), we integrate f''(t) with respect to t.
∫(f''(t)) dt = ∫(4 + 6t) dt
Using the power rule of integration, we have:
∫(f''(t)) dt = 4t + 3t^2 + C1
where C1 is the constant of integration.
Step 2: Find f(t):
To find f(t), we integrate f'(t) with respect to t.
∫(f'(t)) dt = ∫(4t + 3t^2 + C1) dt
Again, using the power rule, we have:
∫(f'(t)) dt = 2t^2 + t^3 + C1t + C2
where C2 is the constant of integration.
So, the function f(t) is given by:
f(t) = 2t^2 + t^3 + C1t + C2
Now, we'll use the initial conditions to determine the values of C1 and C2.
Given f(1) = 3:
Plugging in t = 1 into the equation f(t) = 2t^2 + t^3 + C1t + C2, we get:
3 = 2(1)^2 + (1)^3 + C1(1) + C2
3 = 2 + 1 + C1 + C2
Given f(-1) = -2:
Plugging in t = -1 into the equation f(t) = 2t^2 + t^3 + C1t + C2, we get:
-2 = 2(-1)^2 + (-1)^3 + C1(-1) + C2
-2 = 2 - 1 - C1 + C2
Now, we have two simultaneous equations:
2 + 1 + C1 + C2 = 3
2 - 1 - C1 + C2 = -2
Simplifying the equations, we get:
C1 + C2 = 0
-C1 + C2 = -5
Adding the two equations, we have:
2C2 = -5
Dividing both sides by 2, we find:
C2 = -5/2
Using C2 = -5/2 in the first equation:
C1 + (-5/2) = 0
C1 = 5/2
So, the function f(t) is:
f(t) = 2t^2 + t^3 + (5/2)t - (5/2)
Finally, replacing t with x, our solution is:
f(x) = 2x^2 + x^3 + (5/2)x - (5/2)