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

Question

William Abdalla · Accepted Answer

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!

Isaiah Allen · Answer

undefined To find $ f(x) $ given $ f''(x) = -2 + 36x - 12x^2 $ and $ f(0) = 2, f'(0) = 1 $, follow these steps:

1. Integrate $ f''(x) $ to find $ f'(x) $.
2. Integrate $ f'(x) $ to find $ f(x) $.
3. Apply initial conditions $ f(0) = 2 $ and $ f'(0) = 1 $ to solve for the constants of integration.

Step 1:
$$ \int f''(x) \,dx = \int (-2 + 36x - 12x^2) \,dx $$
$$ f'(x) = -2x + 18x^2 - 4x^3 + C_1 $$

Step 2:
$$ \int f'(x) \,dx = \int (-2x + 18x^2 - 4x^3 + C_1) \,dx $$
$$ f(x) = -x^2 + 6x^3 - x^4 + C_1x + C_2 $$

Step 3:
Apply initial conditions:
$$ f(0) = 2 \Rightarrow 2 = 0 + 0 + 0 + C_2 $$
$$ f'(0) = 1 \Rightarrow 1 = 0 + 0 + 0 + C_1 $$

Therefore, $ C_1 = 1 $ and $ C_2 = 2 $.

So, the solution is:
$$ f(x) = -x^2 + 6x^3 - x^4 + x + 2 $$