0
$\begingroup$

Let $H(x)$ be the cubic Hermite interpolation of $f(x)=x^4+1$ on the interval $[0,1]$ interpolating at $x=0$ and $x=1$. Then

  1. $\max_{x\in I} |f(x)-H(x)|=1/16$.
  2. The maximum of $|f(x)-H(x)|$ is attained at $1/2$.
  3. $\max_{x\in I} |f(x)-H(x)|=1/21.$
  4. The maximum of $|f(x)-H(x)|$ is attained at $1/4$.

I find Hermite polynomial as $2x^3-3x^2+4x-1$ (may be wrong). But my answer did't match with any option. Please suggest me. Thank a lot.

$\endgroup$

1 Answer 1

1
$\begingroup$

First note that $H(x) =2x^3-x^2+1$, so that $f(x)-H(x)=x^2(1-x)^2$, the maximum is attaind at $x=1/2$ and is equal to $1/16$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .