Unlocking the Inverse of Quadratic Function: The Invertible Journey

	Definition of First Function	Substitute Second Function	Simplify
$f (f^{- 1} (x)) = ? x$	$2 f^{- 1} (x) - 3 = ? x$	$2 (\frac{x + 3}{2}) - 3 = ? x$	$x = x ✓$
$f^{- 1} (f (x)) = ? x$	$\frac{f ( x ) + 3}{2} = ? x$	$\frac{2 x - 3 + 3}{2} = ? x$	$x = x ✓$

Definition of First Function

Substitute Second Function

Simplify

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

2 f^{- 1} (x) - 3 = ? x

2 (\frac{x + 3}{2}) - 3 = ? x

x = x ✓

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

\frac{f ( x ) + 3}{2} = ? x

\frac{2 x - 3 + 3}{2} = ? x

x = x ✓

Domain of $f (x)$	Inverse Function
$0 \leq x < \infty$	$f^{- 1} (x) = x$
$- \infty < x \leq 0$	$f^{- 1} (x) = - x$

Domain of

f (x)

Inverse Function

0 \leq x < \infty

f^{- 1} (x) = x

- \infty < x \leq 0

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

Consider the following quadratic function.

f (x) = 4 x^{2} - 9

Solve

y = f (x)

for

x .

What is the positive solution?

Find the negative input when the output is

7 .

We want to solve the equation y=f(x) for x. Let's consider the given function rule. f(x)=4x^2-9 We will now substitute the given expression for f(x) in y=f(x) and then solve for x.

Since we are asked to find the positive solution, the answer is x=sqrt(y+94) or x= sqrt(y+9)2.

Note that the input is represented by the x-variable, and the output is represented by the y-variable. As we found in Part A, two inputs — one positive and one negative — can give us one output. We are asked to find the negative input when the output is 7. To do so, we will use the equation that gives the negative x-values. x=-sqrt(y+9)/2 Now, we will substitute 7 for y into the above equation.

Consider the following graph.

Which graph represents the reflection of $f$ across the line $y = x ?$

We will reflect the graph of f across the line y=x. Let's start by drawing y=x. Since it is already in the slope-intercept form, we can easily identify its slope and y-intercept. y=x ⇔ y= 1x+ 0 Our line will pass through the origin with a slope of 1.

Next, we will reflect the points of the given function across y=x. Remember that when a point (x,y) is reflected across the line y=x, the coordinates of the reflected point will be (y,x). With this in mind, let's make a table to identify the coordinates of five reflected points.

Given Points	Reflected Points
(-2,6)	(6,-2)
(-1,0)	(0,-1)
(0,-2)	(-2,0)
(1,0)	(0,1)
(2,6)	(6,2)

Let's now connect the reflected points in a smooth curve.

By removing the other elements, let's focus solely on the reflected graph g.

According to what we graphed, the correct answer is option D.

Consider the following quadratic function.

f (x) = \frac{1}{2} x^{2} - 6

In which of the intervals is the function invertible?

A . B . C . D . - 6 \leq x < 6 0 \leq x - \frac{1}{2} \leq x \leq 6 - 6 < x \leq \frac{1}{2}

Write the inverse of the function in the interval from Part A.

We want to find an interval on which we can define the inverse of the given function. Since the function is quadratic, we have to restrict the domain to avoid having repeated output values. To do so, let's identify the vertex of the function. cc Quadratic Function& Vertex f(x) = a(x- h)^2+ k & ( h, k) [1em] Quadratic Function & Vertex f(x) = 1/2(x- 0)^2+( -6) & ( 0, -6) We can restrict the domain to values either equal or less than or equal or greater than the x-coordinate of the vertex. x ≤ 0 or 0 ≤ x Notice that any smaller interval included in the above intervals can be considered a solution. However, in the domains that include both positive and negative numbers, our function is not invertible. Therefore, the answer is option B.

We will now find the inverse of our function. To do so, we will first replace f(x) with y. f(x)=1/2x^2-6 ⇔ y=1/2x^2-6 To algebraically determine the inverse of the given relation, we exchange x and y and solve for y. ccc Given Equation& & Inverse Equation y=1/2 x^2-6 & & x=1/2 y^2-6 The result of isolating y in the new equation will be the inverse of the given function.

Let's now use function notation and replace y with g(x). y=± sqrt(2x+12) ⇕ g(x)=± sqrt(2x+12) Notice that we ended with two expressions — one non-negative and one non-positive. Since the domain of f(x) chosen in Part A is 0≤ x, all its input values are non-negative. Inputs of f(x) are the outputs of its inverse g(x). Therefore, the inverse function's rule should be the non-negative expression. g(x)=sqrt(2x+12)

Consider the following graph.

Determine whether the inverse of

f

is a function or not.

We are given the following function and its graph. We want to determine whether the inverse of this function is also a function or not.

We will use the Horizontal Line Test. If the horizontal lines intersect the graph once, then the inverse is also a function. Conversely, if there is even one horizontal line that intersects the graph more than once, then the inverse is not a function. With this in mind, we will draw several horizontal lines across the given graph of a function.

We can see above that there are horizontal lines that intersect the curve at more than one point. Therefore, the inverse of the given function is not a function.

Inverses of Quadratic Functions

Catch-Up and Review

Are Quadratic and Square Root Functions Related?

Inverse Functions

Example

Invertible Function

Condition for Invertibility of a Function

Proof

Restricting the Domain of a Quadratic Function to Guarantee Invertibility

Hint

Solution

Identifying Invertible Functions

Finding the Inverse of a Quadratic Function

Modeling Falling Objects

Answer

Hint

Solution

Practice Finding a Point on the Inverse Function

Analyzing and Predicting the Trajectory of a Projectile

Hint

Solution

The Relation Between Quadratic and Radical Functions

Inverses of Quadratic Functions

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