+0

# Let

+2
87
2
+497

Let

$$f(x) = \begin{cases} k(x) &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}$$

Find the function $k(x)$ such that $f$ is its own inverse.

michaelcai  Sep 24, 2017
Sort:

#1
+79786
+2

y =  2 + ( x - 2)^2

y - 2  =  (x - 2)^2       take both roots

±√[ y - 2 ] = x - 2      add 2 to both sides

±√[ y - 2 ] + 2  =  x    exchange x and y

±√[ x - 2 ] + 2 =  y  =  possible choices for k(x)

Since  +√[ x - 2 ] + 2   > 0    for all x  >2, then  -√[ x - 2 ] + 2  is the correct  function

[ To see this, note that   (-1,11)  is a point on  2 + ( x - 2)^2......but ( 11, -1) is not on  +√[ x - 2 ] + 2 .....however,  (11 , -1)  is on  -√[ x - 2 ] + 2  ]

So....k(x)  =  -√[ x - 2 ] + 2

See  the  inverses here : https://www.desmos.com/calculator/iiz2jtxdok

CPhill  Sep 24, 2017
#2
+91229
+1

Hi Chris,

This question confused me but what you have done is certainly correct. :)

I have included the graph here.

For the benefit of others.

The inverse of a function is the reflection of that function in the line y=x.

That is why I included y=x on my graph.

Melody  Sep 25, 2017

### 6 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details