Answer to Question #156221 in Calculus for Rashinda

Question

Given that f is the real valued function defined by f(x) = 1 / (1 + e^x). And I = (1/4, 1/2).
(i)Show that if x E I, then f(x) E I. Where E represents "an element of".
(ii) Calculate f''(x), hence show that if x E I, thenㅣ f'(x)ㅣ<= 1/4.
(iii) Deduce that if x E I, then ㅣf(x) - sㅣ<= 1/4 ㅣx - sㅣ
s E (1/4, 1/2).

Answer 1

(i) $x\in I, I=(\dfrac{1}{4}, \ \dfrac{1}{2})$

$f(x)=\dfrac{1}{1+e^x}$

$f'(x)=(\dfrac{1}{1+e^x})'=-\dfrac{e^x}{(1+e^x)^2}<0, x\in \R$

Therefore the function $f(x)$ decreases for $x\in \R.$

$f(\dfrac{1}{4})=\dfrac{1}{1+e^{1/4}}<\dfrac{1}{1+(1)^{1/4}}=\dfrac{1}{1+1}=\dfrac{1}{2}$

$f(\dfrac{1}{2})=\dfrac{1}{1+e^{1/2}}>\dfrac{1}{1+(4)^{1/2}}=\dfrac{1}{1+2}=\dfrac{1}{3}>\dfrac{1}{4}$

Therefore

$\dfrac{1}{4}<f(x)<\dfrac{1}{2}, x\in(\dfrac{1}{4},\ \dfrac{1}{2})$

Therefore if $x\in I,$ then $f(x)\in I.$

(ii)

$f'(x)=(\dfrac{1}{1+e^x})'=-\dfrac{e^x}{(1+e^x)^2}<0, x\in \R$

$f''(x)=(-\dfrac{e^x}{(1+e^x)^2})'$

$=-\dfrac{e^x(1+e^x)^2-2e^{2x}(1+e^x)}{(1+e^x)^4}$

$=\dfrac{e^x(e^x-1)}{(1+e^x)^3}>0, x>0$

Therefore for $x>0$ the function $f'(x)$ increases and $f(x)<0.$

Therefore for $x>0$ the function $|f'(x) |$ decreases and $|f(x)|>0.$

$|f'(x)|<|f'(0)|=\dfrac{e^{0}}{(1+e^{0})^2}=\dfrac{1}{4}, x\in I$

$|f'(x)|\leq\dfrac{1}{4}, x\in I$

(iii)

Let $x=0.3, s=0.3$

$f(0.3)=\dfrac{1}{1+e^{0.3}}>\dfrac{1}{1+1.5}=0.4$

$|f(x)-s|>|0.4-0.3|=0.1$

$\dfrac{1}{4}|x-s|=\dfrac{1}{4}|0.3-0.3|=0$

Hence the statement

$|f(x)-s|\leq\dfrac{1}{4}|x-s|, s\in(\dfrac{1}{4}, \ \dfrac{1}{2})$

is False.