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Question Number 38286 by mondodotto@gmail.com last updated on 23/Jun/18

$$\left(\mathit{i}\right)\mathrm{given}\mathrm{the}\mathrm{function}\mathit{f}\left(\mathit{t}\right)={\mathbf{e}}^{\mathit{t}}\mathbf{and}\mathit{g}\left(\mathit{t}\right)=\mathbf{ln}\mathit{t}$$$$\mathbf{show}\mathbf{that}\mathit{f}\circ \mathit{g}\left(\mathit{t}\right)=\mathit{g}\circ \mathit{f}\left(\mathit{t}\right)$$$$\left(\mathit{i}\mathit{i}\right)\mathrm{if}\mathit{f}\left(\mathit{t}\right)=\mathit{a}\mathit{t},\mathit{g}\left(\mathit{t}\right)={\mathit{b}\mathit{t}}^2+3$$$$\left(\mathit{f}\mathit{o}\mathit{g}\right)\left(2\right)=35\mathbf{and}\left(\mathit{f}\mathit{o}\mathit{g}\right)\left(3\right)=75$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathit{a}\mathrm{and}\mathit{b}$$

Commented by math khazana by abdo last updated on 24/Jun/18

$$1)wehave\mathrm{\forall }t\in Rgof\left(t\right)=ln\left(e^t\right)=t$$$$\mathrm{\forall }t\in ]0,+\mathrm{\infty }[fog\left(t\right)=e^{ln\left(t\right)}=tsowehave$$$$fog=gofonlyon]0,+\mathrm{\infty }[!$$$$ii)fog\left(t\right)=f\left(g\left(t\right)\right)=f({bt}^2+3)=a({bt}^2+3)$$$$fog\left(t\right)=f\left(g\left(t\right)\right)=ag\left(t\right)=a({bt}^2+3)$$$$$$$$fog\left(2\right)=35\Rightarrow a(4b+3)=35\Rightarrow 4ab+3a=35$$$$fog\left(3\right)=75\Rightarrow a(9b+3)=75\Rightarrow 9ab+3a=75\Rightarrow$$$$36ab+27a=9.35and36ab+12a=4.75\Rightarrow$$$$15a=9.35-4.75\Rightarrow a=\frac{9.35}{15}-\frac{4.75}{15}$$$$=\frac{3.3.5.7}{3.5}-\frac{4.3.5.5}{3.5}=21-20=1\Rightarrow$$$$4b=35-3=32\Rightarrow b=8atthiscasef\left(t\right)=tand$$$$g\left(t\right)=8t^2+3$$

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