contestada

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

Respuesta :

sqdancefan
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
Ver imagen sqdancefan
lublana

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

[tex]p=t^{-1}(p)[/tex]

Therefore, the given function is inverse function of itself.

Hence, option a is true.

b.y(j)=[tex]-\frac{1}{j}

Let y(j)=y then we get

[tex]y=-\frac{1}{j}[/tex]

[tex]j=-\frac{1}{y}[/tex]

[tex]j=-\frac{1}{y(j)}[/tex]

[tex]j=-\frac{1}{\frac{-1}{j}}[/tex]

[tex]j=j[/tex]

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

c.[tex]w(y)=-\frac{2}{y}[/tex]

Suppose that w(y)=w

Then [tex]w=-\frac{2}{y}[/tex]

[tex]y=-\frac{2}{w}[/tex]

[tex]w(y)=-\frac{2}{-\frac{2}{w}}[/tex]

[tex]w(y)=w[/tex]

[tex]w(y)=-\frac{2}{y}[/tex]

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

d.[tex]d(p)=\frac{1}{x^2}[/tex]

Let d(p)=d

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

[tex]d=\frac{1}{x^2}[/tex]

[tex]x^2=\frac{1}{d}[/tex]

[tex]x=\sqrt{\frac{1}{d}}[/tex]

[tex]x=\sqrt{\frac{1}{d(p)}[/tex]

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

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