# Thread: Help with lagrange problem

1. ## Help with lagrange problem

The third degree Taylor polynomial of f around x=2 is...

T(x) = 7 - 9(x-2)^2 - 3(x-2)^3

The fourth derivative satisfies the inequality [abs(f''''(x))] <_ 6 for all x in the closed interval [0,2]. Use the Lagrange error bound on the approximation of f(0) via T(0) to explain why f(0) is negative.

2. Originally Posted by Kaitosan
The third degree Taylor polynomial of f around x=2 is...

T(x) = 7 - 9(x-2)^2 - 3(x-2)^3

The fourth derivative satisfies the inequality [abs(f''''(x))] <_ 6 for all x in the closed interval [0,2]. Use the Lagrange error bound on the approximation of f(0) via T(0) to explain why f(0) is negative.
$\displaystyle T(0) = 7 - 36 + 24 = -5$

error = $\displaystyle |f(0) - T(0)| \leq \left|\frac{6(-2)^4}{4!}\right| = 4$

therefore,

$\displaystyle -4 \leq f(0) - T(0) \leq 4$

$\displaystyle -4 \leq f(0) - (-5) \leq 4$

$\displaystyle -4 \leq f(0) + 5 \leq 4$

$\displaystyle -9 \leq f(0) \leq -1$

3. **applauds**

Thank you!

As a sidenote, I think you were supposed to input x=2 as your "c" but it wouldn't have made a difference anyways. Right?

4. Originally Posted by Kaitosan
**applauds**

Thank you!

As a sidenote, I think you were supposed to input x=2 as your "c" but it wouldn't have made a difference anyways. Right?
you've got the wrong idea ...

"c" is where the Taylor polynomial is centered, not the function input ... the function input is x = 0.

$\displaystyle T(x) = 7 - 9(x-2)^2 - 3(x-2)^3$ ... c = 2

$\displaystyle T(0) = 7 - 9(0-2)^2 - 3(0-2)^3$

the Lagrange error bound is $\displaystyle \frac{M(x-c)^{n+1}}{(n+1)!}$

... $\displaystyle \frac{6(0-2)^4}{4!}$

5. Yeah, that's right. I mis-worded, sorry.

I've a question. "C" is supposed to be a number between "a" (the number that the series is centered around) and "x" (the actual output of the function) that maximizes the Lagrange error.... right?

6. Originally Posted by Kaitosan
Yeah, that's right. I mis-worded, sorry.

I've a question. "C" is supposed to be a number between "a" (the number that the series is centered around) and "x" (the actual output of the function) that maximizes the Lagrange error.... right?
notation differs ... check your textbook.