Find the cubic polynomial f(x) such that

Question

Find the cubic polynomial f(x) such that

$(−1)=-4$, $'(−1)=4$, $''(−1)=-6$, $'''(−1)=12$

I know that i have to use an augmented matrix in order to solve it, so i decided to use the formula $a+bx+cx^2+dx^3$. But after that, do i have to input every x in the a,b,c,d and created the augmented matrix?

Answer 1

So let us use $f(x)=ax^3+bx^2+cx+d$
$\therefore f'(x)=3ax^2+2bx+c$
$\therefore f''(x)=6ax+2b$
$\therefore f'''(x)=6a$

$f'''(-1)=6a=12 \implies a=2$
$f''(-1)=-12+2b=-6\implies b=3$
$f'(-1)=6-6+c=4\implies c=4$
$f(-1)=-2+3-4+d=-4\implies d=-1$

$$\therefore f(x)=2x^3+3x^2+4x-1$$

Answer 2

Guide:

Let $$f(x) = a + b(x+1) + c(x+1)^2 + d(x+1)^3$$

$$f(-1) = a $$

Let $$f'(x) = b + 2c(x+1) + 3d(x+1)^2$$

$$f'(-1) = b $$

We have $$f(x) = \sum_{i=0}^3\frac{f^{(i)}(-1)}{i!}(x+1)^i$$

Answer 3

When all the equations involve the same $x$ value it is much easier. You have $f(x)=a+bx+cx^2+dx^3$. Then $f'''(x)=6d$ so $d=2$. Now evaluate $f''(-1)$. As you know $d$ it is easy and gives $c$. Two more steps and you have the cubic.