Bri S.

asked • 07/30/20# Consider the function f ( x ) whose second derivative is f ' ' ( x ) = 2 x + 2 sin ( x ) . If f ( 0 ) = 3 and f ' ( 0 ) = 3 , what is f ( 3 ) ?

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

this is my work

f''(x)=2x+2sin(x)

f'(x)=2x^2/2+2sinx+c

f'(x)=x^2+2sinx+c

f'(0)=0^2+2cos(0)+c=3

0+2(1)+c=3

c=1---> f'(x)=x^2+cosx+x

f(x)=x^3/3+sinx+x^2/2+c

f(0)=0^3/3+sin(0)+(0)^2/2=3

c=3

f(x)=x^3/3+sinx+x^2/2+3

so f(3)=3^3/3+sin(3)+3^2/2/+3= 16.55 which is wrong.. Where did I go wrong?

## 1 Expert Answer

Richard P. answered • 07/30/20

PhD in Physics with 10+ years tutoring experience in STEM subjects

In getting from f'' to t there are two constants of integration ( I will call them A and B)

f(x) = A + B x + (1/3) x^{3} - 2 sin(x)

f' = B + x^{2} - 2 cos(x)

f'' = 2x + 2 sin(x)

From the initial conditions 3 = A and 3 = B-2

So the final form of f is f(x) = 3 + 5 x + (1/3) x^{2} - 2 sin(x)

So f(3) = 21 - 2 sin(3)

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Mike D.

07/30/20