Bri S.

# 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?

Mike D.

tutor
Bri. Your idea to find f'(x) was correct - integrating f''(x) but when you integrate sin x you should get -cos x , so you made a mistake there. So f'(x) should be x^2 - 2 cos x + c. f'(0) = 3 so 0-2+c=3, giving c=5. Try and continue from here ...
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07/30/20