If f′′(x)=9e3x+4cos2x, f(0)=3 and f′(0)=13, find f(x).

Question

If f′′(x)=9e3x+4cos2x, f(0)=3 and f′(0)=13, what is f(x).

Answer 1

f '(x) = ∫f ''(x) dx

So f '(x) = ∫(9e^3x + 4cos(2x))dx = 3e^3x + 2sin(2x) + C

Since f′(0) = 13 then:

13 = 3e³⁽⁰⁾ + 2sin(2•0) + C

13 = 3(1) + 2(0) + C

C = 10

So f '(x) = 3e^3x + 2sin(2x) + 10

Integrating again to get f(x):

f(x) = ∫f '(x) dx = ∫(3e^3x + 2sin(2x) + 10)dx = e^3x - cos(2x) + 10x + C

Since f(0) = 3 we can say:

3 = e³⁽⁰⁾ - cos(2•0) + 10(0) + C

3 = 1 - 1 + 0 + C

C = 3

So f(x) = e^3x - cos(2x) + 10x + 3