Question

Find the function f, if: f'(x)=2/(x^3) + 4e^x + 5, f(-1)=1, f(1)=-1 (Note: Consider the domain and write the answer in ascending order of the variable).

Answer 1

f(x) = -1/x^2 + 4e^x + 5x + 1 + 1/x - 4e^(-1)

Explanation:

We can find the function f(x) by integrating f'(x) with respect to x:

â«f'(x) dx = â«(2/(x^3) + 4e^x + 5) dx

f(x) = -1/x^2 + 4e^x + 5x + C

To find the constant C, we can use the given initial conditions:

f(-1) = 1 = -1/(-1)^2 + 4e^(-1) - 5 + C

C = 1 + 1/1 - 4e^(-1)

f(x) = -1/x^2 + 4e^x + 5x + 1 + 1/x - 4e^(-1)

Therefore, the function f(x) is:

f(x) = -1/x^2 + 4e^x + 5x + 1 + 1/x - 4e^(-1)