if-g-x-f-x-f-1-x-and-f-2-x-lt-0-then-show-that-g-x-is-increasing-in-0-1-2-and-g-x-is-decreasing-in-1-2-1-

Question Number 27204 by rish@bh last updated on 03/Jan/18

if g(x)=f(x)+f(1−x) and f^((2)) (x)<0 then show that g(x) is increasing in (0,1/2) and g(x) is decreasing in (1/2,1)

$$\mathrm{if}\:{g}\left({x}\right)={f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right) \\ $$$$\mathrm{and}\:{f}^{\left(\mathrm{2}\right)} \left({x}\right)<\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that}\: \\ $$$${g}\left({x}\right)\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{in}\:\left(\mathrm{0},\mathrm{1}/\mathrm{2}\right)\:\mathrm{and} \\ $$$${g}\left({x}\right)\:\mathrm{is}\:\mathrm{decreasing}\:\mathrm{in}\:\left(\mathrm{1}/\mathrm{2},\mathrm{1}\right) \\ $$

Answered by Giannibo last updated on 03/Jan/18

∃g′ because ∃f′′⇒∃f′ g′(x)=f′(x)−f′(1−x) f′′(x)<0 ⇒f′↓ If x<1−x ⇔x<(1/2) ⇒^(f′↓) f′(x)>f′(1−x) g′(x)>0⇒ g↑ If x>1−x⇔x>(1/2) ⇒^(f′↓) f′(x)<f′(1−x) g′(x)<0 ⇒ g↓

$$ \\ $$$$ \\ $$$$\exists\mathrm{g}'\:\mathrm{because}\:\exists\mathrm{f}''\Rightarrow\exists\mathrm{f}' \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\mathrm{f}'\left(\mathrm{x}\right)−\mathrm{f}'\left(\mathrm{1}−\mathrm{x}\right) \\ $$$$\mathrm{f}''\left(\mathrm{x}\right)<\mathrm{0}\:\Rightarrow\mathrm{f}'\downarrow \\ $$$$\mathrm{If}\:\mathrm{x}<\mathrm{1}−\mathrm{x}\:\Leftrightarrow\mathrm{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\overset{\mathrm{f}'\downarrow} {\Rightarrow}\mathrm{f}'\left(\mathrm{x}\right)>\mathrm{f}'\left(\mathrm{1}−\mathrm{x}\right) \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)>\mathrm{0}\Rightarrow\:\mathrm{g}\uparrow \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{x}>\mathrm{1}−\mathrm{x}\Leftrightarrow\mathrm{x}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\overset{\mathrm{f}'\downarrow} {\Rightarrow}\mathrm{f}'\left(\mathrm{x}\right)<\mathrm{f}'\left(\mathrm{1}−\mathrm{x}\right) \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)<\mathrm{0}\:\Rightarrow\:\mathrm{g}\downarrow \\ $$

Commented by rish@bh last updated on 03/Jan/18