Suppose $f$ is a twice differentiable function such that $\underset {a \le x \le b} {\sup} |f''(x)|=M_2$. Then the maximum error in linear interpolation of $f$ on $[a,b]$ is given by
$(a)$ $\frac {hM_2} {8}.$
$(b)$ $\frac {h^2 M_2} {8}.$
$(c)$ $\frac {h^2 M_2} {2}.$
$(d)$ $\frac {h^2 M_2} {6}.$
We know that if $g(x)$ is the linear interpolating polynomial of $f$ on $[a,b]$ then we would have $g(x) =f(a) + (x-a) \frac {\{f(b)-f(a)\}} {b-a}$. Now the error in linear interpolation is given by $E(x)=f(x) - g(x) =f(x) - \{f(a) + (x-a) \frac {\{f(b)-f(a)\}} {b-a} \}$. Since $f$ is twice differentiable $\exists$ $\xi \in (a,b)$ such that $\frac {f(b)-f(a)} {b-a}=f'(\xi).$ So we have $E(x)=f(x)-f(a)-(x-a)f'(\xi).$ Again applying Lagranges mean value theorem to $f$ on $[a,x]$ we have $\exists$ $\eta \in (a,x)$ such that $f(x)-f(a)=(x-a)f'(\eta).$ So we have $E(x)=(x-a)(f'(\eta)-f'(\xi)).$ Without loss of generality let us assume that $\eta \ge \xi.$ Then applying Langranges mean value theorem to $f'$ on $[\xi,\eta]$ we have a point $\zeta \in (\xi,\eta)$ such that $E(x)=(x-a)(\eta-\xi)f''(\zeta).$ Hence we have $|E(x)| \le (b-a)^2 M_2$ for all $x \in [a,b].$ Now since we are talking about linear interpolation the number of nodes are only $2$ which are precisely $a$ and $b$ and so $b-a=h$. Hence we have the maximum value of the error as $h^2M_2.$ Which does not meet my purpose.
Would anybody please give me some suggestions?
Thank you in advance.
HINT. Maybe your estimation is a bit rough. I think this attempt helps you: let $g(x)$-linear interpolating polynomial for twice differentiable function $f$ on $[a;b]$, then for $a \leqslant x \leqslant b$, $$f(x)-g(x)=\tfrac{(x-a)(x-b)}{2}\cdot f''(c)$$ Then, use that: $$\underset{a \leqslant x \leqslant b}{max}(x-a)(b-x)\leqslant \frac{h^2}{4}$$ For more details, see: error in LI