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Question Number 165949 by mathlove last updated on 10/Feb/22

f(x)=((5x−2)/a)  ∧f^(−1) (x)=((x+b)/5)  faind   a×b=?

$${f}\left({x}\right)=\frac{\mathrm{5}{x}−\mathrm{2}}{{a}}\:\:\wedge{f}^{−\mathrm{1}} \left({x}\right)=\frac{{x}+{b}}{\mathrm{5}} \\ $$$${faind}\:\:\:{a}×{b}=? \\ $$

Answered by qaz last updated on 10/Feb/22

(((5x−2)/a))′∙(((x+b)/5))′=(1/a)=1  ⇒a=1  ⇒f(x)=5x−2  f^(−1) (x)=((2+x)/5)=((x+b)/5)  ⇒b=2  ⇒a×b=2

$$\left(\frac{\mathrm{5x}−\mathrm{2}}{\mathrm{a}}\right)'\centerdot\left(\frac{\mathrm{x}+\mathrm{b}}{\mathrm{5}}\right)'=\frac{\mathrm{1}}{\mathrm{a}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{5x}−\mathrm{2} \\ $$$$\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{2}+\mathrm{x}}{\mathrm{5}}=\frac{\mathrm{x}+\mathrm{b}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{a}×\mathrm{b}=\mathrm{2} \\ $$

Answered by cortano1 last updated on 10/Feb/22

 f(x)= ((5x−2)/a)⇔ f^(−1) (x)=((−ax−2)/(−5))=((ax+2)/5)   so  { ((a=1)),((b=2)) :}⇒a×b=2

$$\:\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{5x}−\mathrm{2}}{\mathrm{a}}\Leftrightarrow\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{−\mathrm{ax}−\mathrm{2}}{−\mathrm{5}}=\frac{\mathrm{ax}+\mathrm{2}}{\mathrm{5}} \\ $$$$\:\mathrm{so}\:\begin{cases}{\mathrm{a}=\mathrm{1}}\\{\mathrm{b}=\mathrm{2}}\end{cases}\Rightarrow\mathrm{a}×\mathrm{b}=\mathrm{2} \\ $$

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