How do you find f given #f''(x)=-2+36x-12x^2# and f(0)=2, f'(0)=1?

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Answer 1 · 2016-11-22T20:58:15

We integrate once and then integrate again. We know that #int(x^n)dx = x^(n + 1)/(n + 1) + C#.

#int(-12x^2 + 36x - 2)dx = -4x^3 + 18x^2 - 2x + C#

That's #f'(x)#, so when #x = 0#, #y = 1# inside this function.

#1 = -4(0)^3 + 18(0)^2 - 2(0) + C#

#1 = C#

The first derivative is hence #f'(x) = -4x^3 + 18x^2 - 2x + 1#

We integrate this:

#int(-4x^3 + 18x^2 - 2x + 1) = -x^4 + 6x^3 - x^2 + x + C#

That's #f(x)#, so when #x = 0#, #y= 2# inside this function.

#2 = -0^4 + 6(0)^3 - 0^2 + 0 + C#

#2 = C#

The original function is therefore #f(x) = -x^4 + 6x^3 - x^2 + x + 2#

{Check by finding #(d^2y)/(dx^2)#}.

Hopefully this helps!