Consider the function

Question

Consider the function f(x) whose second derivative is  f''(x)=4x+4sin(x). If f(0)=2 and f'(0)=2, what is f(3)?

Answer 1

Given f"(x) = 4x + 4sinx x

f'(x) = 2x^2-4cos x + C the constant of integration

Since f'(0) = 2

Then

2 =2(0)^2 - 4 cos(0) +C

2 = -4(1) +C

6= C

Now f'(x) = 2x^2 - 4 cos x + 6

This makes

f(x) = (2/3) x^3 -4 six x + 6x + D (D stands for constant of integration since C used already)

Since f(0) = 2 we get

2 = (2/3) 0*3 -4 sin 0 + 6(0) + D

2 = D

Thus

f(x) = (2/3) x^3 -4 six x + 6x + 2

f(3) = (2/3)*3^3 - 4 sin 3 + 18 + 2

f(3) = 18 -4 sin 3 + 20

f(3) = 38 - 4 sin 3

Answer 2

We must take the integral twice of f'' to find f and then use that expression to find f(3).

1. The integral of 4x+sin(x) is f'(x) = 2x² - 4cos(x) + c

1. Now plug in the point f'(0)=2 to find C
2. 2=0-4cos(0)+c --> c=2+4=6
3. The expression for f'(x) = 2x² - 4cos(x) + 6

2. The integral of 2x² - 4cos(x) + 6 is f(x) = 2x³/3 - 4sin(x) + 6x + c

1. Now plug in the point f(0)=2 to find C
2. 2=0-4sin(0)+0+c --> c=2+0=2
3. The expression for f(x) = 2x³/3 - 4sin(x) + 6x + 2

3. Now use the expression that we found for f(x) and plug in x=3

1. f(3) = 2(3)³/3 - 4sin(3) + 6(3) + 2
2. f(3) = 38 - 4sin(3)