Consider the function f(x) whose second derivative is f″(x)=6x+7sin(x). If f(0)=3 and f′(0)=3, what is f(x)? f(x)= ?

Question

Consider the function f(x) whose second derivative is f″(x)=6x+7sin(x).

If f(0)=3 and f′(0)=3, what is f(x)?

f(x)= ?

Answer 1

In order to find f(x), one must first find f'(x); therefore, one must first integrate f''(x):

1. f'(x) = ∫ f''(x) dx
2. f'(x) = ∫[6x + 7 sin(x)] dx
3. f'(x) = ∫6x dx + ∫7 sin(x) dx
4. f'(x) = 6 ∫x dx + 7 ∫sin(x) dx
5. f'(x) = (6/2)x² - 7 cos(x) + C
6. f'(x) = 3x² - 7 cos(x) + C

Now, the constant C must be determined. The problem states that f'(0) = 3; substituting 0 for x and 3 for f'(x):

3x² - 7 cos(x) + C = f'(x)

3(0)² - 7 cos(0) + C = 3

0 - 7 (1) + C = 3; where cos(0) = 1

0 - 7 + C = 3

C = 3 + 7

C = 10

Therefore:

f'(x) = 3x² - 7 cos(x) + 10

Now, one must find f(x); therefore, one must integrate f'(x):

1. f(x) = ∫ f'(x) dx
2. f(x) = ∫ (3x² - 7 cos(x) - 4) dx
3. f(x) = ∫3x² dx - ∫ 7 cos(x) dx - ∫4 dx
4. f(x) = (3/3)x³ - 7 ∫ cos(x) dx - 4 ∫dx
5. f(x) = x³ - 7 sin(x) -4x + C

Now, again, one must determine the constant C. The problem states that f(0) = 3; substituting 0 for x and 3 for f(x):

x³ - 7 sin(x) - 4x + C = f(x)

(0)³ - 7 sin(0) - 4(0) + C = 3

0 - 0 - 0 + C = 3

C = 3

The final equation for the function f(x):

f(x) = x³ - 7 sin(x) - 4x + 3

Answer 2

f '(x) = ∫f "(x)dx and f(x) = ∫f '(x)dx

f '(x) = 3x²+7cos(x) + C, f '(0) = 3 = 0 + 7 + C --> C = -4 --> f '(x) = 3x²+7cos(x) -4

f(x) = x³-7sin(x) -4x + C, f(0) = 0 - 0 - 0 + C = 3 or C = 3

f(x) = x³-7sin(x) -4x + 3