# `f(x) = -3 + lnx` Find: (a) `f^(-1)` (Inverse of f) (b) domain and range of `f^(-1)` (c) graph of `f^(-1)`

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`f(x)=-3+ lnx`

(a) To determine the inverse of the given function, we need to replace f(x) with y.

`y=-3 + lnx`

Then, interchange x and y.

`x=-3 + ln y`

From here, let's isolate y.

`x+3=lny`

Express the equation into its equivalent exponent form. Note that the exponent form of `ln a = m` is `e^m=a` .

`y =e^(x+3)`

Then, replace y with `f^(-1)(x)` .

**Hence, the inverse of the given function is `f^(-1)(x) = e^(x+3)` .**

(b) Note that there are no values of x that would result to undefined values of y in the inverse function `y=e^(x+3)` .

**Hence, domain of `f^(-1)` is all real numbers. In interval notation, domain is `(-oo,+oo)` .**

To determine the range of `y=e^(x+3)` , let's consider the properties of basic exponential function `y=e^x` .

The range for this basic function is:

`e^xgt0`

So to solve for the range of `y=e^(x+3)` , multiply both sides of the given properties by `e^3` .

`e^x*e^3 gt 0*e^3`

`e^x*e^3gt0`

`e^(x+3)gt0`

Then, replace `e^(x+3)` with y.

`ygt0`

**Hence, the range of `f^(-1)` is greater than zero. In interval notation, range is `(0,+oo)` .**

(c) So the graph of `f^(-1)(x) = e^(x+3)` is: