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