Question

Help me please! (:
Consider the function f(x) whose second derivative is f''(x) = 8 x + 5 \sin (x). If f(0) = 2 and f'(0) = 3, what is f(x)?

Answer 1

well u need to integrate it twice :) \[f''(x) = 8 x + 5 \sin x\]\[\int f''(x) \ \text{d}x=\int (8 x + 5 \sin x) \ \text{d}x\]\[f'(x)=4x^2-5\cos x +c\]one of given conditions \(f'(0)=3\) put this in the last expression\[f'(0)=3=-5+c\]\[c=8\]finally\[f'(x)=4x^2 -5 \cos x+8\]i think u got it from here...what will be f(x) ??? :) :)

Answer 2

how would I underive 4x^2? @mukushla

Answer 3

b/c it would be __x^3+5sinx+C

Answer 4

anti derivative of \(x^n\) is\[\frac{x^{n+1}}{n+1}\]and be careful :) \[\int \sin x \ \text{d} x=-\cos x+c\]\[\int \cos x \ \text{d} x=\sin x+c\]

Answer 5

\[\frac{ 4 }{ 3 }x^3 -5sinx\] ?

Answer 6

@mukushla

Answer 7

u dropped integral of 8 !! try again :)

Answer 8

\[\frac{ 4 }{ 3 }x^3-5sinx+C\]

Answer 9

well we found that\[f'(x)=4x^2 -5 \cos x+8\]then integrating\[\int f'(x) \text{d} x=\int \ (4x^2 -5 \cos x+\color\red{8}) \ \text{d} x\]\[f(x)=\frac{ 4 }{ 3 }x^3-5 \sin x+8x+c\]am i right?

Answer 10

Ohhhhh... I see lol. no we evaluate at f(0)=2? right?

Answer 11

\[f(0)=2=8+C\]

Answer 12

very right :)

Answer 13

oh lol wait\[f(0)=2=0+c\]

Answer 14

\[f(x)=\frac{ 4 }{ 3 }x^3-fsinx+8x-6\]

Answer 15

0+C ??

Answer 16

look at it once again :) more carefully

Answer 17

how did you get f'(0)=3?

Answer 18

thats a given condition in problem

Answer 19

That I understand. but where did the -5+C come from? @mukushla

Answer 20

ahh i got\[f'(x)=4x^2-5\cos x+c\]from integration, now let \(x=0\)\[f'(0)=4\times 0-5 \cos 0 +c=-5+c\]

Answer 21

ohhhh.... haha..
\[f(x)=\frac{ 4 }{ 3 }x^3-5sinx+8x+0\]
@mukushla

Answer 22

well c=2 and final answer will be\[f(x)=\frac{ 4 }{ 3 }x^3-5\sin x+8x+2\]check it again :)