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Question Number 188568 by 073 last updated on 03/Mar/23

Answered by floor(10²Eta[1]) last updated on 03/Mar/23

$$\mathrm{f}\left(\mathrm{h}(-4)\right)$$$$3^{-\mathrm{x}}+3^{\mathrm{x}}=-4\Rightarrow 1+3^{2\mathrm{x}}=-4.3^{\mathrm{x}}$$$$\Rightarrow 3^{2\mathrm{x}}+4.3^{\mathrm{x}}+1=0\Rightarrow 3^{\mathrm{x}}=\sqrt{3}-2\Rightarrow \mathrm{x}={\log }_3(\sqrt{3}-2)$$$$\Rightarrow \mathrm{h}(-4)=9^{-{\log }_3(\sqrt{3}-2)}+9^{{\log }_3(\sqrt{3}-2)}-2$$$$=\frac{1}{{(\sqrt{3}-2)}^2}+{(\sqrt{3}-2)}^2-2=\frac{1}{7-4\sqrt{3}}+5-4\sqrt{3}$$$$=\frac{84-48\sqrt{3}}{7-4\sqrt{3}}=12$$$$\Rightarrow \mathrm{f}\left(\mathrm{h}(-4)\right)=\mathrm{f}\left(12\right)$$$${\mathrm{x}}^3+4=12\Rightarrow \mathrm{x}=2\Rightarrow \mathrm{f}\left(12\right)=-7$$$$$$$$\mathrm{f}\left(\mathrm{h}(-4)\right)=-7$$

Commented by 073 last updated on 03/Mar/23

$$\mathrm{nice}\mathrm{solution}$$$$\mathrm{thanks}$$

Answered by manxsol last updated on 04/Mar/23

$$y=3^x+\frac{1}{3^x}\Rightarrow y^2=9^x+\frac{1}{9^x}+2$$$$h\left(y\right)=y^2-4$$$$h\left(x\right)=x^2-4$$$$foh$$$$\left(\begin{array}{c}\mathrm{\Re }\\ \end{array}\right)\stackrel{\curvearrowright }{h}\left(\begin{array}{c}[-4,\stackrel{}{+\mathrm{\infty }}]\\ \end{array}\right)\stackrel{\curvearrowright }{f}\left(\begin{array}{c}[-7,+\mathrm{\infty }]\\ \end{array}\right)$$$$f\left(h(-4)\right)=$$$$h(-4)=12$$$$f\left(12\right)=f(x^3+4)=1-4x$$$$12=x^3+4x\Rightarrow x=2$$$$f\left(12\right)=1-4\left(2\right)$$$$f\left(12\right)=-7$$$$f\left(h(-4)\right)=-7$$

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