Question

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

Answer 1

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

Answer 2

Answer:

a,b and c.

Step-by-step 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)=$-\frac{1}{j}Let y(j)=y then we get [tex]y=-\frac{1}{j}$

$j=-\frac{1}{y}$

$j=-\frac{1}{y(j)}$

$j=-\frac{1}{\frac{-1}{j}}$

$j=j$

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

c.$w(y)=-\frac{2}{y}$

Suppose that w(y)=w

Then $w=-\frac{2}{y}$

$y=-\frac{2}{w}$

$w(y)=-\frac{2}{-\frac{2}{w}}$

$w(y)=w$

$w(y)=-\frac{2}{y}$

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

d.$d(p)=\frac{1}{x^2}$

Let d(p)=d

If we replace $\frac{1}{x^2}by p then we get [tex]d=\frac{1}{x^2}$

$x^2=\frac{1}{d}$

$x=\sqrt{\frac{1}{d}}$

$x=\sqrt{\frac{1}{d(p)}$

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