Question

Which of the following functions are their own inverses? Select all that apply.

a. t(p) = p
b. y(j) = -1/j
c. w(y) = -2/y
d. d(p) = 1/x^2

Answer 1

a,b and c.

Explanation:

We have to find the the functions that are their own inverses.

a.t(p)=p

Then the inverse function of given function is

`p=t^(-1)(p)`

Therefore, the given function is inverse function of itself.

Hence, option a is true.

b.y(j)=
`-(1)/(j)</p><p>Let y(j)=y then we get </p><p>[tex]y=-(1)/(j)`

`j=-(1)/(y)`

`j=-(1)/(y(j))`

`j=-(1)/((-1)/(j))`

`j=j`

Hence, the function is inverse of itself.Therefore, option b is true.

c.
`w(y)=-(2)/(y)`

Suppose that w(y)=w

Then
`w=-(2)/(y)`

`y=-(2)/(w)`

`w(y)=-(2)/(-(2)/(w))`

`w(y)=w`

`w(y)=-(2)/(y)`

Hence, the function is inverse function of itself.Therefore, option c is true.

d.
`d(p)=(1)/(x^2)`

Let d(p)=d

If we replace
`(1)/(x^2)by p then we get </p><p>[tex]d=(1)/(x^2)`

`x^2=(1)/(d)`

`x=\sqrt{(1)/(d)}`

`x=\sqrt{(1)/(d(p))`

Hence, the function is not self inverse function.Therefore, option d is false.

Answer 2

A function is its own inverse if it is symmetrical about the line y=x. This is the case for functions t, y, w. Function d(x) = 1/x^2 is symmetrical about the line x=0, but is not symmetrical about the line y=x.

The appropriate choices are ...
a. t(p) = p
b. y(j) = -1/j
c. w(y) = -2/y