1. ## higher-order derivatives

find a quadratic function f such that f(3)=33, f'(3)=22, f''(3)=8

2. Originally Posted by skeske1234
find a quadratic function f such that f(3)=33, f'(3)=22, f''(3)=8
start with $\displaystyle f(x) = ax^2 + bx + c$

since $\displaystyle f(3) = 33$ ... $\displaystyle 33 = a(9) + b(3) + c$

$\displaystyle f'(x) = 2ax + b$

since $\displaystyle f'(3) = 22$ ... $\displaystyle 22 = 2a(3) + b$

$\displaystyle f''(x) = 2a$

since $\displaystyle f''(3) = 8$ ... $\displaystyle 8 = 2a$

so ... find $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$.

3. Originally Posted by skeske1234
find a quadratic function f such that f(3)=33, f'(3)=22, f''(3)=8
Assume $\displaystyle f(x)=ax^2+bx+c$

Now, apply the given conditions and determine a,b,c.

4. Or you can use the Taylor series of the polynomial about $\displaystyle x=3$, and not have to solve for any unknowns:

$\displaystyle f(x)=f(3)+\frac{f'(3)}{1!}(x-3)+\frac{f''(3)}{2!}(x-3)^2 =33+22(x-3)+4(x-3)^2 = 4x^2-2x+3$