finding y = f(x)

**dasani** · July 1st 2007, 10:19 AM

Lost on this problem:

Find y = f(x) given f^ii(x) = 12x^2 and f^i (1) = 10 and f(2) = 26

**Jhevon** · July 1st 2007, 10:29 AM

Originally Posted by dasani

Lost on this problem:

Find y = f(x) given f^ii(x) = 12x^2 and f^i (1) = 10 and f(2) = 26

How long did you try to figure this out?

$f''(x) = 12x^2$

$\Rightarrow f'(x) = \int f''(x)~dx = 12 \int x^2 ~dx = 4x^3 + C$

Now $f'(1) = 10$

$\Rightarrow 10 = 4(1)^3 + C \implies C = 6$

So, $f'(x) = 4x^3 + 6$

$\Rightarrow f(x) = \int f'(x)~dx = \int \left( 4x^3 + 6 \right)~dx = x^4 + 6x + C$

Now $f(2) = 26$

$\Rightarrow 26 = (2)^4 + 6(2) + C \implies C = -2$

So, finally, $\boxed { f(x) = x^4 + 6x - 2 }$

**red_dog** · July 1st 2007, 10:29 AM

$\displaystyle f''(x)=12x^2\Rightarrow f'(x)=12\cdot \frac{x^3}{3}+c_1=4x^3+c_1$
$f'(1)=10\Rightarrow 4+c_1=10\Rightarrow c_1=6\Rightarrow f'(x)=4x^3+6$
$\displaystyle \Rightarrow f(x)=4\cdot \frac{x^4}{4}+6x+c_2=x^4+6x+c_2$
$f(2)=26\Rightarrow c_2=-2\Rightarrow f(x)=x^4+6x-2$

**dasani** · July 1st 2007, 10:32 AM

I've been working on these problems since yesterday...We haven't covered this chapter in class so I'm basically teaching myself, which is why I am so confused about it.

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