Find the function f where f ''(x)=8x³+5, f(1)=0 and f '(1)=8.
f' = 2 x^4 + 5x + c ... 2 + 5 + c = 8 ... c = 1
f = 2/5 x^5 + 5/2 x^2 + x + c ... 2/5 + 5/2 + 1 + c = 0 ... c = -39/10
Thanks! This helps a lot.
To find the function f, we will integrate the given second derivative f ''(x) and solve for the constant of integration using the initial conditions f(1) = 0 and f'(1) = 8.
First, let's integrate the second derivative f ''(x) = 8x³ + 5 to find the first derivative f '(x):
f '(x) = ∫(8x³ + 5) dx
= 8 ∫(x³) dx + 5 ∫(1) dx
= 8 ∫(x³) dx + 5x
= (8/4)x^4 + 5x + C₁
= 2x^4 + 5x + C₁
Next, let's find the constant of integration C₁ by using the condition f'(1) = 8:
f '(1) = 2(1)^4 + 5(1) + C₁
= 2 + 5 + C₁
= 7 + C₁
Since f'(1) = 8, we can set 7 + C₁ equal to 8 and solve for C₁:
7 + C₁ = 8
C₁ = 8 - 7
C₁ = 1
Now that we have the first derivative f '(x) = 2x^4 + 5x + 1, we can integrate it to find the original function f(x):
f(x) = ∫(2x^4 + 5x + 1) dx
= (2/5)x^5 + (5/2)x^2 + x + C₂
Finally, we will use the condition f(1) = 0 to solve for the constant of integration C₂:
f(1) = (2/5)(1)^5 + (5/2)(1)^2 + 1 + C₂
= (2/5) + (5/2) + 1 + C₂
= 2/5 + 5/2 + 1 + C₂
= 2/5 + 10/5 + 5/5 + C₂
= 17/5 + C₂
Since f(1) = 0, we can set 17/5 + C₂ equal to 0 and solve for C₂:
17/5 + C₂ = 0
C₂ = -17/5
Therefore, the function f(x) with f ''(x) = 8x³ + 5, f(1) = 0, and f'(1) = 8 is:
f(x) = (2/5)x^5 + (5/2)x^2 + x - 17/5