Let $f$ be a continuous function on $[a,b]$, and twice differentiable on $(a,b)$. Let $T(x)$ be the local linearization of $f$ at $x=a$. Show that the error in the linear approximation, $|T(x)−f(x)|$ for $x \in (a,b)$ is no bigger than $\kappa(b−a)^2$ , where $|f′′(x)|≤\kappa$ for all $x \in (a,b)$. (That is, $\kappa$ is an upper bound for the size of the concavity of $f$ on the interval.)
I have been struggling with this, would really be grateful if you showed your working as well.
The Taylor series expansion of $f(x)$ at $a$ is $$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+...+\frac{f^{n}(a)}{n!}(x-a)^2+R_n$$ where $R_n=\frac{f^{n+1}(\zeta)}{(n+1)!}(x-a)^{n+1}$ is the trancation error when the series is truncated at the $n$th derivative, and $\zeta$ lies between $a$ and $x$.
$T(x)$ is a first-order linear approximation. Hence $$T(x)=f(a)+f'(a)(x-a)$$ The truncation error is $$R_n=\frac{f''(\zeta)}{2!}(x-a)^2$$ But since $\zeta\in(a,\, x)$ and $x\in[a,\,b]$ and $|f''(x)|<\kappa\Rightarrow -\kappa<f''(x)<\kappa$ in this range, $$|f(x)-T(x)|<\frac{2\kappa}{2}(b-a)^2$$