higher-order derivatives

**skeske1234** · August 8th 2009, 04:59 PM

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

**skeeter** · August 8th 2009, 05:05 PM

Originally Posted by skeske1234

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

start with $f(x) = ax^2 + bx + c$

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

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

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

$f''(x) = 2a$

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

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

**malaygoel** · August 8th 2009, 05:05 PM

Originally Posted by skeske1234

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

Assume $f(x)=ax^2+bx+c$

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

**Bruno J.** · August 8th 2009, 06:05 PM

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

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

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