Answer to Question #103882 in Calculus for BIVEK SAH

Question

Use Lagrange’s mean value theorem to prove that


x-x^2/2《log(1+x)《x-x^2/2(1+x) if x》0.

Answer 1

$x-\frac{x^2}{2}\leq\log (1+x)\leq x-\frac{x^2}{2(1+x)}, \ \ \quad \forall \ x\geq0$

Proof:

Lagrange's Mean Value Theorem:

Let $f: [a,b]\rightarrow \mathbb{R}$ be a continuous function, differentiable on the open interval $(a,b)$  . Then there exists some  $c\in(a,b) \$ such that $f^\prime(c)=\frac{f(b)-f(a)}{b-a}$ .

$f(x)=\log(1+x)+\frac{x^2}{2}$

This function is continuous in $[0,b]$ , differentiable in $(0,b),$ for all $b\geq0.$

Using Lagrange’s Mean Value Theorem for function $f(x)$, we have:

$\exists c\in (0,b): \ \frac{f(b)-f(0)}{b-0}=f^\prime(c)$

$\frac{f(b)-f(0)}{b-0}= \frac{(\log(1+b)+\frac{b^2}{2})-(\log(1)+0)}{b}= \frac{\log(1+b)+\frac{b^2}{2}}{b}=f^\prime(c)=\frac{1}{1+c}+c$

$\frac{\log(1+b)+\frac{b^2}{2}}{b}=\frac{1}{1+c}+c=1+\frac{c^2}{1+c}\geq1 \Rightarrow \frac{\log(1+b)+\frac{b^2}{2}}{b}\geq1 \Rightarrow \log(1+b)+\frac{b^2}{2}\geq b$

$\Rightarrow \log(1+b)\geq b-\frac{b^2}{2}, \forall b\geq0$

So, $x-\frac{x^2}{2}\leq\log (1+x), \ \ \forall \ x\geq 0$

$f(x)=\log(1+x) +\frac{x^2}{2(1+x)}$

This function is continuous in $[0,b]$ , differentiable in $(a,b)$ for all $b\geq 0.$

Using Lagrange’s Mean Value Theorem for function $f(x)$ , we have:

$\exists c\in (0,b): \ \frac{f(b)-f(0)}{b-0}=f^\prime(c)$

$\frac{f(b)-f(0)}{b-0}= \frac{(\log(1+b)+\frac{b^2}{2(1+b)})-(\log(1)+0)}{b}= \frac{\log(1+b)+\frac{b^2}{2(1+b)}}{b}=f^\prime(c)=$

$=\frac{1}{1+c}+\frac{2c\times 2(1+c)-2c^2}{4(1+c)^2}=\frac{2c^2+8c+4}{4c^2+8c+4}=1-\frac{2c^2}{4c^2+8c+4} \leq 1$

$\frac{\log(1+b)+\frac{b^2}{2(1+b)}}{b}\leq 1\Rightarrow {\log(1+b)+\frac{b^2}{2(1+b)}}\leq b \Rightarrow {\log(1+b)\leq b- \frac{b^2}{2(1+b)}}, \ \forall b\geq 0$

So, $\log (1+x)\leq x-\frac{x^2}{2(1+x)}, \ \ \quad \forall \ x\geq0$