Answer: f(x) = -sin(x) + (9/2)x^2 + 5x + 8
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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
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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
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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
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We have
f(x) = -sin(x) + (9/2)x^2 + 5x + E
turn into
f(x) = -sin(x) + (9/2)x^2 + 5x + 8