Question

Consider the functions f(x) = 2x and g(x) =
`(1)/(x-3)`≠ 3

(a) Calculate (f ° g)(4)

(b) Find
`g^(-1) (x)`

(c) Write doen the domain of
`g^(-1)`

Answer 1

Answer to Q1:

(fog)(4) = 2

Explanation:

We have given two function. We have to find their composition.

f(x) = 2x and g(x)= 1 / x-3

(fog)(x) = ? and (fog)(4) = ?

The formula to find composition is:

(fog)(x) = f(g(x))

(fog)(x) = f(1 / x-3)

(fog)(x) = 2(1 / x-3)

(fog)(x) = 2 / x-3

Putting x = 4 in above equation, we have

(fog)(4) = 2 / 4-3

(fog)(4) = 2 / 1

(fog)(4) = 2

Answer to Q2:

g⁻¹(x) = 1/x+3

Explanation:

We have given a function and we have to find its inverse.

g(x) = 1 / x-3

g⁻¹(x) = ?

Let y = g(x)

y = 1 / x-3

We have to separate x from above equation.

y(x-3) = 1

x-3 = 1 / y

Adding 3 to both sides of above equation, we have

x-3+3 = 1/y+3

x = 1/y+3

Putting x = g⁻¹(y) in above equation, we have

g⁻¹(y) = 1/y+3

Replacing y with x , we have

g⁻¹(x) = 1/x+3 which is the answer.

Answer to Q3:

(-∞,0)∪(0,∞)

Explanation:

Since g⁻¹(x) = 1/x+3

We have to find the domain of above function.

Domain is defined as the set of values of independent variable where function is defined.

Hence given function contain 1/x term which is defined all real values except at x = 0.

The term 3 is defined at all real values.

Hence,g⁻¹(x) has domain equal to all real values except x = 0.

dom g⁻¹(x) = (-∞,0)∪(0,∞).

Answer 2

Question 1:

For this case we must find
`(f_ {0} g) (4)`knowing that:

`f (x) = 2x\\g (x) = \frac {1} {x-3}`

By definition we have that, be two functions f (x) and g (x), then the composite function of f with g is:

`(g_ {0} f) (x) = g [f (x)]`

In this case, they ask us for the function composed of g with f:

`(f_ {0} g) (x) = f [g (x)]`

So, we have:

`(f_ {0} g) (x) = 2 (\frac {1} {x-3})\\(f_ {0} g) (4) = 2 (\frac {1} {4-3})\\(f_ {0} g) (4) = 1`

ANswer:

`(f_ {0} g) (4) = 1`

Question 2:

For this case, we must find the inverse function of
`g (x) = \frac {1} {x-3}`, given by:
`g ^ {- 1} (x)`

To do this, replace g(x) with y:

`y = \frac {1} {x-3}`

We exchange variables:

`x = \frac {1} {y-3}`

We solve for "y":

`y-3 = \frac {1} {x}\\y = \frac {1} {x} +3`

Replace "y" with
`g ^ {-1} (x)`

So, we have:

`g ^ {- 1} (x) = \frac {1} {x} +3`

Answer:

`g ^ {- 1} (x) = \frac {1} {x} +3`

Question 3:

For this case, we have by definition, the domain of a function f (x), is the set of all the values ​​for which the function is defined.

We must find the domain of the following function:

`g ^ {- 1} (x) = \frac {1} {x} +3`

It is observed that the function is not defined for
`x = 0`

Then the domain is given by all the values ​​of x, except 0.

`x \\eq 0`for any integer n

Answer:

All numbers, except
`x = 0`