Question Number 38286 by mondodotto@gmail.com last updated on 23/Jun/18
$$\left(\boldsymbol{{i}}\right)\:\mathrm{given}\:\mathrm{the}\:\mathrm{function}\:\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)=\boldsymbol{\mathrm{e}}^{\boldsymbol{{t}}} \:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{g}}\left(\boldsymbol{{t}}\right)=\boldsymbol{\mathrm{ln}{t}} \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{{f}}\circ\boldsymbol{{g}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{g}}\circ\boldsymbol{{f}}\left(\boldsymbol{{t}}\right) \\ $$$$\left(\boldsymbol{{ii}}\right)\mathrm{if}\:\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{at}}\:,\:\boldsymbol{{g}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{bt}}^{\mathrm{2}} +\mathrm{3} \\ $$$$\left(\boldsymbol{{fog}}\right)\left(\mathrm{2}\right)=\mathrm{35}\:\boldsymbol{\mathrm{and}}\:\left(\boldsymbol{{fog}}\right)\left(\mathrm{3}\right)=\mathrm{75} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\boldsymbol{{a}}\:\mathrm{and}\:\boldsymbol{{b}} \\ $$
Commented by math khazana by abdo last updated on 24/Jun/18
![1) we have ∀t ∈R gof(t)=ln(e^t )=t ∀t∈]0,+∞[ fog(t)=e^(ln(t)) =t so we have fog=gof only on]0,+∞[ ! ii)fog(t)=f(g(t))=f(bt^2 +3)=a(bt^2 +3) fog(t)=f(g(t))=ag(t)=a(bt^2 +3) fog(2)=35⇒a(4b+3)=35 ⇒4ab +3a=35 fog(3)=75 ⇒a(9b +3)=75 ⇒9ab +3a=75 ⇒ 36ab +27 a= 9.35 and 36ab +12a=4.75 ⇒ 15a =9.35 −4.75⇒a = ((9.35)/(15)) −((4.75)/(15)) = ((3.3.5.7)/(3.5)) −((4.3.5.5)/(3.5)) =21 −20=1 ⇒ 4b=35−3=32 ⇒b=8 at this case f(t)=t and g(t)=8t^2 +3](https://www.tinkutara.com/question/Q38324.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\forall{t}\:\in{R}\:{gof}\left({t}\right)={ln}\left({e}^{{t}} \right)={t} \\ $$$$\left.\forall{t}\in\right]\mathrm{0},+\infty\left[\:{fog}\left({t}\right)={e}^{{ln}\left({t}\right)} ={t}\:{so}\:{we}\:{have}\:\right. \\ $$$$\left.{fog}={gof}\:{only}\:{on}\right]\mathrm{0},+\infty\left[\:!\right. \\ $$$$\left.{ii}\right){fog}\left({t}\right)={f}\left({g}\left({t}\right)\right)={f}\left({bt}^{\mathrm{2}} +\mathrm{3}\right)={a}\left({bt}^{\mathrm{2}} +\mathrm{3}\right) \\ $$$${fog}\left({t}\right)={f}\left({g}\left({t}\right)\right)={ag}\left({t}\right)={a}\left({bt}^{\mathrm{2}} \:+\mathrm{3}\right) \\ $$$$ \\ $$$${fog}\left(\mathrm{2}\right)=\mathrm{35}\Rightarrow{a}\left(\mathrm{4}{b}+\mathrm{3}\right)=\mathrm{35}\:\Rightarrow\mathrm{4}{ab}\:+\mathrm{3}{a}=\mathrm{35} \\ $$$${fog}\left(\mathrm{3}\right)=\mathrm{75}\:\Rightarrow{a}\left(\mathrm{9}{b}\:+\mathrm{3}\right)=\mathrm{75}\:\Rightarrow\mathrm{9}{ab}\:+\mathrm{3}{a}=\mathrm{75}\:\Rightarrow \\ $$$$\mathrm{36}{ab}\:+\mathrm{27}\:{a}=\:\mathrm{9}.\mathrm{35}\:{and}\:\mathrm{36}{ab}\:+\mathrm{12}{a}=\mathrm{4}.\mathrm{75}\:\Rightarrow \\ $$$$\mathrm{15}{a}\:=\mathrm{9}.\mathrm{35}\:−\mathrm{4}.\mathrm{75}\Rightarrow{a}\:=\:\frac{\mathrm{9}.\mathrm{35}}{\mathrm{15}}\:−\frac{\mathrm{4}.\mathrm{75}}{\mathrm{15}} \\ $$$$=\:\frac{\mathrm{3}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{\mathrm{3}.\mathrm{5}}\:−\frac{\mathrm{4}.\mathrm{3}.\mathrm{5}.\mathrm{5}}{\mathrm{3}.\mathrm{5}}\:=\mathrm{21}\:−\mathrm{20}=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{4}{b}=\mathrm{35}−\mathrm{3}=\mathrm{32}\:\Rightarrow{b}=\mathrm{8}\:{at}\:{this}\:{case}\:{f}\left({t}\right)={t}\:{and} \\ $$$${g}\left({t}\right)=\mathrm{8}{t}^{\mathrm{2}} \:+\mathrm{3} \\ $$