Answer:
a.
![f^(-1)(x)=(x+3)/(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/x5bj1vqmzprzqbccfp7rn8y6nrll6v5cns.png)
b.
![g^(-1)(x) =e^{(x)/(2)}+1](https://img.qammunity.org/2020/formulas/mathematics/high-school/382p8tgsy8asbcj7uonsju5ni81vds9jlj.png)
c.
![h^(-1)(x)=(e^(y))/(e^(y)-1)](https://img.qammunity.org/2020/formulas/mathematics/high-school/kmmhf22rw4ksleh4mydd0f760kxn4x0yh8.png)
d.
![k^(-1)(x)=-(2log(5-y))/(log(3))](https://img.qammunity.org/2020/formulas/mathematics/high-school/r3t23ahep8915o10ccvo357t8el701aj2s.png)
Explanation:
Here is the procediment for each case.
a.
as we know
then
finding the expresion of x, we find the inverse of the function, then:
![y=2x-3\\y+3=2x\\\\(y+3)/(2)=x](https://img.qammunity.org/2020/formulas/mathematics/high-school/ki040b6ubbpri7wdz880mu1k8d8m2r8afu.png)
Then:
![f^(-1)(x)=(x+3)/(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/x5bj1vqmzprzqbccfp7rn8y6nrll6v5cns.png)
b.
for this we have to remember that the inverse function of the log is the exp, then:
![g(x) = 2 log(x-1)\\y=2log(x-1)\\\\\\(y)/(2)=log(x-1)\\\\e^{(y)/(2)}=e^(log(x-1))\\\\e^{(y)/(2)}=x-1\\e^{(y)/(2)}+1=x](https://img.qammunity.org/2020/formulas/mathematics/high-school/6m0cp5495iz9na2pj2tuu2dg46vkxrhs9c.png)
Then the inverse function is:
![g^(-1)(x) =e^{(x)/(2)}+1](https://img.qammunity.org/2020/formulas/mathematics/high-school/382p8tgsy8asbcj7uonsju5ni81vds9jlj.png)
c. In this case we have to also remember the relation between the e and the ln, then:
with the properties of the ln we have:
now finding the inverse function we have:
![h(x)=ln\left((x)/(x-1)\right)\\\\y=ln\left((x)/(x-1)\right)\\\\e^(y)=e^{ln\left((x)/(x-1)\right)}\\\\\\e^(y)=(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=(e^(y))/(e^(y)-1)](https://img.qammunity.org/2020/formulas/mathematics/high-school/epux0ay40o1mslja1itehcob6993nbsldk.png)
then:
![h^(-1)(x)=(e^(y))/(e^(y)-1)](https://img.qammunity.org/2020/formulas/mathematics/high-school/kmmhf22rw4ksleh4mydd0f760kxn4x0yh8.png)
d. in the last one we have:
then:
![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)=-(x)/(2)log(3)\\-2log(5-y)=xlog(3)\\-(2log(5-y))/(log(3))=x](https://img.qammunity.org/2020/formulas/mathematics/high-school/t9r9i4r7ps6v1nfmnfouyrxjsm3p7mscb5.png)
then:
![k^(-1)(x)=-(2log(5-y))/(log(3))](https://img.qammunity.org/2020/formulas/mathematics/high-school/r3t23ahep8915o10ccvo357t8el701aj2s.png)