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Question Number 107454 by Rio Michael last updated on 10/Aug/20

Given the function f(x) = ((x + 3)/(x−2)) and g(x) = (1/2)xe^x (1) Find the centre of symmetry of f. (2) Define the monotony of g and if possible draw a variation table for g(x). (3) Sketch the function g(x) (4) determine if f and g intersect.

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:=\:\frac{{x}\:+\:\mathrm{3}}{{x}−\mathrm{2}}\:\mathrm{and}\:\mathrm{g}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}{xe}^{{x}} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{symmetry}\:\mathrm{of}\:{f}. \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Define}\:\mathrm{the}\:\mathrm{monotony}\:\mathrm{of}\:\mathrm{g}\:\mathrm{and}\:\mathrm{if}\:\mathrm{possible}\:\mathrm{draw}\:\mathrm{a}\:\mathrm{variation} \\ $$$$\mathrm{table}\:\mathrm{for}\:\mathrm{g}\left({x}\right). \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Sketch}\:\mathrm{the}\:\mathrm{function}\:\mathrm{g}\left({x}\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{determine}\:\mathrm{if}\:{f}\:\mathrm{and}\:\mathrm{g}\:\mathrm{intersect}. \\ $$

Answered by abdomsup last updated on 11/Aug/20

f(x)=((x+3)/(x−2)) =((x−2+5)/(x−2)) =1+(5/(x−2)) ⇒f(x)−1 =(5/(x−2)) let w(2,1) let provethat f(4−x) =2−f(x) (f(2a−x)=2b−f(x)) f(4−x)=((4−x+3)/(4−x−2)) =((7−x)/(2−x))=((x−7)/(x−2)) 2−f(x)=2−((x+3)/(x−2)) =((2x−4−x−3)/(x−2)) =((x−7)/(x−2)) ⇒f(4−x)=2−f(x) ⇒ w(2,1) is a centre of symetry

$${f}\left({x}\right)=\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\:=\frac{{x}−\mathrm{2}+\mathrm{5}}{{x}−\mathrm{2}}\:=\mathrm{1}+\frac{\mathrm{5}}{{x}−\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)−\mathrm{1}\:=\frac{\mathrm{5}}{{x}−\mathrm{2}}\:{let}\:{w}\left(\mathrm{2},\mathrm{1}\right) \\ $$$${let}\:{provethat}\:{f}\left(\mathrm{4}−{x}\right)\:=\mathrm{2}−{f}\left({x}\right) \\ $$$$\left({f}\left(\mathrm{2}{a}−{x}\right)=\mathrm{2}{b}−{f}\left({x}\right)\right) \\ $$$${f}\left(\mathrm{4}−{x}\right)=\frac{\mathrm{4}−{x}+\mathrm{3}}{\mathrm{4}−{x}−\mathrm{2}}\:=\frac{\mathrm{7}−{x}}{\mathrm{2}−{x}}=\frac{{x}−\mathrm{7}}{{x}−\mathrm{2}} \\ $$$$\mathrm{2}−{f}\left({x}\right)=\mathrm{2}−\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\:=\frac{\mathrm{2}{x}−\mathrm{4}−{x}−\mathrm{3}}{{x}−\mathrm{2}} \\ $$$$=\frac{{x}−\mathrm{7}}{{x}−\mathrm{2}}\:\Rightarrow{f}\left(\mathrm{4}−{x}\right)=\mathrm{2}−{f}\left({x}\right)\:\Rightarrow \\ $$$${w}\left(\mathrm{2},\mathrm{1}\right)\:{is}\:{a}\:{centre}\:{of}\:{symetry} \\ $$$$ \\ $$

Commented by Rio Michael last updated on 11/Aug/20

Thank you so much sir.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir}. \\ $$

Answered by abdomsup last updated on 11/Aug/20

2)g(x)=(1/2)x e^x lim_(x→−∞) g(x)=0 lim_(x→+∞) g(x)=+∞ g^′ (x) =(1/2)e^x +(1/2)xe^x =(1/2)(x+1)e^x g^′ (x)≥0 ⇒x≥−1 vsruation x −∞ −1 +∞ g^′ − + g 0 decr −(1/(2e)) incr +∞

$$\left.\mathrm{2}\right){g}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{x}\:{e}^{{x}} \\ $$$${lim}_{{x}\rightarrow−\infty} {g}\left({x}\right)=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow+\infty} {g}\left({x}\right)=+\infty \\ $$$${g}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{e}^{{x}} \:+\frac{\mathrm{1}}{\mathrm{2}}{xe}^{{x}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}\right){e}^{{x}} \\ $$$${g}^{'} \left({x}\right)\geqslant\mathrm{0}\:\Rightarrow{x}\geqslant−\mathrm{1}\:\:{vsruation} \\ $$$${x}\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${g}^{'} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$${g}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:{decr}\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}{e}}\:\:\:{incr}\:\:\:\:\:+\infty \\ $$$$ \\ $$

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