Question

Let f(x) = VE-2(a) What is the domain off?Domain of f =(b) What is the range of f?Range of f =(c) What is the domain off-Domain off-1 =(G) What is the range of f-"?Range of f-1 =(e) Find the formula for f-'(z):

Answer 1

We have the next function

`f(x)=\sqrt[]{x-2}`

The domain is the set of all possible values that x can have in this case we need to remember that we can have a negative value for the radical therefore the domain is

`\: x\ge\: 2`

In interval notation

`\: \lbrack2,\: \infty\: )`

a) [2,inf)

Then for the range is the set of all the possible values that the function can have in this case the range is

`\: \: f(x)\ge0`

In interval notation

`\: \lbrack0,\: \infty\: )`

b) [0,inf)

Then we need to find the inverse of the function given, we make f(x)=y

`y=\sqrt[]{x-2}`

Then we make x=y and y=x

`x=\sqrt[]{y-2}`

Then we isolate the y

`y=x^2+2`

The domain of this function is

`\: \mleft(-\infty\: ,\: \infty\: \mright)`

c) (-inf,inf)

The range of this function is

`\: \lbrack2,\: \infty\: )`

d) [2,inf)

Then as we calculate before the inverse function of the function given is

`f^(-1)(x)=x^2+2`

e) f^-1(x)=x^2+2

ANSWER

a) domain of f: [2,inf)

b) range of f: [0,inf)

c) domain of f^-1: (-inf,inf)

d) range of f^-1: [2,inf)

e) f^-1(x)=x^2+2

`f^(-1)(x)=x^2+2`