Find f. f '''(x) = cos(x), f(0) = 8, f '(0) = 4, f ''(0) = 9 f(x) =

Work Shown:

f ''' (x) = cos(x) third derivative

f '' (x) = sin(x)+C ... integrate both sides to get second derivative. Don't forget the +C at the end

We are given f '' (0) = 9, so we'll make use of this to find C

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

f '' (0) = sin(0)+C

9 = sin(0) + C

9 = 0 + C

9 = C

C = 9

Therefore, f '' (x) = sin(x)+C turns into f '' (x) = sin(x)+9

Integrate both sides of the second derivative to get the first derivative function

f '' (x) = sin(x)+9

f ' (x) = -cos(x)+9x+D ... D is some constant

Make use of f ' (0) = 4 to find D

f ' (x) = -cos(x)+9x+D

f ' (0) = -cos(0)+9(0)+D

4 = -1 + 0 + D

D = 5

So we have f ' (x) = -cos(x)+9x+D turn into f ' (x) = -cos(x)+9x+5

Lastly, apply another round of integrals and substitutions to find the f(x) function. We'll use f(0) = 8.

f ' (x) = -cos(x)+9x+5

f(x) = -sin(x) + (9/2)x^2 + 5x + E E is some constant

f(0) = -sin(0) + (9/2)(0)^2 + 5(0) + E

8 = 0 + 0 + 0 = E

E = 8

We have

f(x) = -sin(x) + (9/2)x^2 + 5x + E

turn into

f(x) = -sin(x) + (9/2)x^2 + 5x + 8

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