Respuesta :
Answer:
a. [tex]f^{-1}(x)=\dfrac{x+3}{2}[/tex]
b. [tex]g^{-1}(x) =e^{\frac{x}{2}}+1[/tex]
c. [tex]h^{-1}(x)=\dfrac{e^{y}}{e^{y}-1}[/tex]
d. [tex]k^{-1}(x)=-\dfrac{2log(5-y)}{log(3)}[/tex]
Step-by-step explanation:
Here is the procediment for each case.
a. [tex]f(x)=2x-3[/tex] as we know [tex]f(x)=y[/tex] then [tex]y=2x-3[/tex] finding the expresion of x, we find the inverse of the function, then:
[tex]y=2x-3\\y+3=2x\\\\\dfrac{y+3}{2}=x[/tex]
Then:
[tex]f^{-1}(x)=\dfrac{x+3}{2}[/tex]
b. [tex]g(x)=2log(x-1)[/tex] for this we have to remember that the inverse function of the log is the exp, then:
[tex]g(x) = 2 log(x-1)\\y=2log(x-1)\\\\\\\dfrac{y}{2}=log(x-1)\\\\e^{\frac{y}{2}}=e^{log(x-1)}\\\\e^{\frac{y}{2}}=x-1\\e^{\frac{y}{2}}+1=x[/tex]
Then the inverse function is:
[tex]g^{-1}(x) =e^{\frac{x}{2}}+1[/tex]
c. In this case we have to also remember the relation between the e and the ln, then:
[tex]h(x) = ln(x)-ln(x-1)[/tex] with the properties of the ln we have:
[tex]h(x) = ln(x)-ln(x-1)=ln\left(\dfrac{x}{x-1}\right)[/tex] now finding the inverse function we have:
[tex]h(x)=ln\left(\dfrac{x}{x-1}\right)\\\\y=ln\left(\dfrac{x}{x-1}\right)\\\\e^{y}=e^{ln\left(\dfrac{x}{x-1}\right)}\\\\\\e^{y}=\dfrac{x}{x-1}}\\\\e^{y}(x-1)=x\\e^{y}x-e^{y}=x\\e^{y}x-x=e^{y}\\x(e^{y}-1)=e^{y}\\x=\dfrac{e^{y}}{e^{y}-1}[/tex]
then:
[tex]h^{-1}(x)=\dfrac{e^{y}}{e^{y}-1}[/tex]
d. in the last one we have: [tex]k(x) =5-3^{-x/2}[/tex] then:
[tex]k(x) =5-3^{-x/2}\\y =5-3^{-x/2}\\y-5=-3^{-x/2}\\5-y=3^{-x/2}\\log(5-y)=log(3^{-x/2})\\log(5-y)=-\frac{x}{2}log(3)\\-2log(5-y)=xlog(3)\\-\dfrac{2log(5-y)}{log(3)}=x[/tex]
then:
[tex]k^{-1}(x)=-\dfrac{2log(5-y)}{log(3)}[/tex]