f ''(x) = 20x3 + 12x2 + 10, f(0) = 7, f(1) = 9
To find the function f(x) using the given information, we can integrate the second derivative of f(x), which is given as f''(x) = 20x^3 + 12x^2 + 10, twice.
The first integration will yield the first derivative f'(x), and the second integration will give us the original function f(x).
Step 1: Integrate f''(x) to find f'(x)
∫f''(x) dx = ∫(20x^3 + 12x^2 + 10) dx
To integrate 20x^3, we use the power rule, which states that ∫x^n dx = (1/(n+1)) * x^(n+1).
∫20x^3 dx = (20/4) * x^4 = 5x^4
To integrate 12x^2, we use the power rule again.
∫12x^2 dx = (12/3) * x^3 = 4x^3
The integral of 10 with respect to x is simply 10x.
∫10 dx = 10x
Adding these three integrals together, we get:
f'(x) = 5x^4 + 4x^3 + 10x + C, where C is the constant of integration.
Step 2: Integrate f'(x) to find f(x)
∫f'(x) dx = ∫(5x^4 + 4x^3 + 10x + C) dx
Integrating each term with respect to x:
∫5x^4 dx = (5/5) * x^5 = x^5
∫4x^3 dx = (4/4) * x^4 = x^4
∫10x dx = 10/2 * x^2 = 5x^2
∫C dx = Cx
Adding these four integrals together, we get:
f(x) = x^5 + x^4 + 5x^2 + Cx + D, where C and D are constants of integration.
Now we will use the given information to find the specific values for C and D.
Given:
f(0) = 7, f(1) = 9
Plugging x = 0 into the function f(x), we get:
f(0) = 0^5 + 0^4 + 5(0^2) + C(0) + D = D
Since f(0) = 7, we know that D = 7.
Plugging x = 1 into the function f(x), we get:
f(1) = 1^5 + 1^4 + 5(1^2) + C(1) + 7 = 9
1 + 1 + 5 + C + 7 = 9
C + 14 = 9
C = 9 - 14
C = -5
Finally, substituting the values of C and D back into the function, we have:
f(x) = x^5 + x^4 + 5x^2 - 5x + 7.
I will assume you want f(x)
f ''(x) = 20x3 + 12x2 + 10
then f'(x) = 5x^4 + 3x^4 + 10x + c
f(x) = x^5 + (3/5)x^4 + 5x^2 + cx + d
f(0) = 7 ---> 7 = 0+0+0+0+d ----------> d = 7
then f(x) = x^5 + (3/5)x^4 + 5x^2 + cx + 7
f(1) = 9 ---> 9 = 1 + 3/5 + 5 + c + 7
c = -23/5
f(x) = x^4 + (3/5)x^4 + 5x^2 - (23/5)x + 7