Find a portion of the domain where the function is one-to-one and find an inverse function.

Question

The function ƒ(x) = −x² + 10 is not one-to-one. Find a portion of the domain where the function is one-to-one and find an inverse function. 

Comments

whats the answer

Answer 1

Metohod to find f is one - to - one finction : Table method :

Make the table of values to find ordered pairs that satisfy the equation.

Choose values for x and find the corresponding values for y.

x	f(x) = - x2 + 10	(x, y)
- 3	f(x) = - (- 3)2 + 10 = 1	(- 3, 1)
- 1	f(x) = - (- 1)2 + 10 = 9	(- 1, 9)
0	f(x) = - (0)2 + 10 = 10	(0, 10)
1	f(x) = - (1)2 + 10 = 9	(1, 9)
3	f(x) = - (3)2 + 10 = 1	(3, 1)

x	y
1	- 3
9	- 1
10	0
9	1
1	3

 

 

 

 

 

 

 

The table on the left is a table of values for f(x) = - x² + 10. The table of values on the right is made up by interchanging the columns of the first table. The table on the right does not represent a function, because the input x = 1 and x = 9 is matched with two different outputs: y = - 3 ,y = 3 and y = - 1, y = 1.

So, f(x) = - x² + 10 is not one-to-one. So, we can' t find a portion of the domain, where the function is one - to - one.

And it does not have an inverse functiion.

Answer 2

Finding an Inverse Function :
1. Use the Horizontal Line Test to decide whether f has an inverse function.
2. In the equation for f(x), replace f(x) by y.
3. Interchange the roles of x and y and solve for y.
4. Replace y by f ^{- 1}(x) in the new equation.
5. Verify that f and f ^{- 1} are inverse functions of each other by showing that the domain of f is equal to the range of f ^{- 1}, the range of f is equal to the domain of  f ^{- 1}, and f ( f ^{- 1}(x) ) = x and f ^{- 1}( f(x) ) = x.

The function is f(x) = - x² + 9.

(1). Applying the Horizontal Line Test :

The graph of the function given by f(x) = - x² + 10 is shown in the above Figure. Because it is possible to find a horizontal line that intersects the graph of f at more than one point, you can conclude that f is not a one-to-one function and does not have an inverse function.

Finding Inverse Functions Informally :

Find the inverse function of f(x) = - x² + 10. Then verify that both f ( f ^{- 1}(x) ) and f ^{- 1}( f(x) ) are equal to the identity function.

The original function f(x) = - x² + 10.

Replace f(x) by y.

y = - x² + 10

Interchange x and y.

x = - y² + 10

Solve for y.

x  - 10 = - y²

- x  + 10 = y²

y = ± √(- x  + 10).

Replace y by f ^{- 1}(x).

f ^{- 1}(x) = ± √(- x  + 10).

The ± indicates that to a given value of x there correspond two values of y. So, y is not a function of x.

The function f(x) = - x² + 10.

The above function is quadratic function, so the domain of the function is all real numbers.

Domain : x ∈ R, where R is real number.

Case 1 : If the domain is x ≤ 10 then f ^{- 1}(x) = + √(- x + 10).

f ( f ^{- 1}(x) ) = f [ √(- x + 10) ] = - [ √(- x + 10) ]² + 10 = - (- x + 10) + 10 = x.

f ^{- 1}( f(x) ) = f ^{- 1} [ - x² + 10 ] = sqrt [ - (- x² + 10) + 10 ] = sqrt [ x² - 10 + 10 ] = x.

f ( f ^{- 1}(x) ) = f ^{- 1}( f(x) ) = x.

Therefore domain x ≤ 10, then inverse function is f ^{- 1}(x) = √(- x + 10).

Case 2 : If the domain is x ≥ 10 then f ^{- 1}(x) = √(- x + 10) does not exist.

For example x = 11 then f ^{- 1}(x) = √(- x + 10) = √[- (11) + 10] = √(- 1) = i(imaginary value).

 

Answer 3

The function is f(x) = - x² + 10.

The above function is quadratic function, so the domain of the function is { x ∈ R }, where R is real number.

The function f(x) = - x² + 10 is symmetrical about y - axis but any horizontal line crosses the graph at two points, so it is one to one function. Either left or right side will be one to one function about y - axis.

Here f(x) = - x² + 10 (x ≥ 0) considered and the inverse function is f ^{- 1}(x) = + √(- x + 10) (x ≤ 10)

The inverse function f ^{- 1}(x) = + √(- x + 10) is symmetrical about x - axis but domain of the original function is (x ≥ 0) the upper portion will be the invesre function as shown in the figure.

 

The graphs of f (x)  and f ^{- 1}(x) are reflections of each other in the line y = x.  we can further verify this reflective property by testing a few points on each graph.

Note that if the point (a, b) = (10, 0) is on the graph of f and the point (b, a) = (0, 10) is on the graph of f ^{- 1.}

 

I hope it helps a little.

Comments

Thank you :) Now, got the answer.

So, if we consider the function left hand side, then the lower part of the inverse function to be considered (f ^{- 1}(x) = - √(- x + 10) (x ≤ 10)) while plotting. Correct me if i am wrong.