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There are many applications of quadratic functions in real life. For example, the height of a projectile changes according to the time squared. A model using a quadratic function receives a time value $t$ and returns the height $h$ of the projectile at that time.
*inverse* of a quadratic function. This lesson will begin by describing the relation between quadratic and square root functions. Then, how and why these functions can define inverses for each other will be understood. ### Catch-Up and Review

What if a person wants to know the time it takes the projectile to reach a specific height? The function could be used to set up a quadratic equation to find that time. However, if many predictions were needed, it would not be practical to solve multiple equations. It would be very convenient to have a function that takes the height and returns the corresponding time.

That convenient function is the

**Here are a few recommended readings before getting started with this lesson.**

Explore

The following applet shows two graphs, one representing the quadratic function $f(x)=x_{2}$ and the other is the square root function $g(x)=x .$ In the applet, both functions have their domains defined as values of $x$ greater than or equal to zero.

Can one graph be derived from the other? If so, how? Based on this exploration, is there any particular relationship between quadratic and square root functions?

Discussion

Every function has an inverse relation. If this inverse relation is also a function, then it is called an inverse function. In other words, the inverse of a function $f$ is another function $f_{-1}$ such that they *undo* each other.

$f(f_{-1}(x))=xandf_{-1}(f(x))=x$

Also, if $x$ is the input of a function $f$ and $y$ its corresponding output, then $y$ is the input of $f_{-1}$ and $x$ its corresponding output.

$f(x)=y⇔f_{-1}(y)=x$

$f(x)=2x−3andf_{-1}(x)=2x+3 $

These functions will be shown to $f(f_{-1}(x))=?x⇕2f_{-1}(x)−3=?x $

Now, in the above equation, $2x+3 $ will be substituted for $f_{-1}(x).$
$2f_{-1}(x)−3=?x$

Substitute

$f_{-1}(x)=2x+3 $

$2(2x+3 )−3=?x$

▼

Simplify left-hand side

DenomMultFracToNumber

$2⋅2a =a$

$(x+3)−3=?x$

RemovePar

Remove parentheses

$x+3−3=?x$

SubTerm

Subtract term

$x=x✓$

Definition of First Function | Substitute Second Function | Simplify | |
---|---|---|---|

$f(f_{-1}(x))=?x$ | $2f_{-1}(x)−3=?x$ | $2(2x+3 )−3=?x$ | $x=x✓$ |

$f_{-1}(f(x))=?x$ | $2f(x)+3 =?x$ | $22x−3+3 =?x$ | $x=x✓$ |

Therefore, $f$ and $f_{-1}$ *undo* each other. The graphs of these functions are each other's reflection across the line $y=x.$ This means that the points on the graph of $f_{-1}$ are the *reversed* points on the graph of $f.$

$f(x)=x_{2} $

The inverse operation of raising a value to the second power is taking its square root. In other words, squaring a number and calculating its square root are operations that undo each other. The following example demonstrates this concept.
$8_{2}=6464 =8 $

For this reason, square root functions can be used to define the inverse of quadratic functions. There is something else to consider, however. There is another possibility for the previous example.
$(-8)_{2}=6464 =8 $

The fact that the square of two different numbers can have the same value complicates defining an inverse for a quadratic function. In this case, the existence of an inverse operation is not enough for a function to be invertible. Before exploring the conditions a function must meet to be invertible, the definition of an invertible function will be given. Concept

A function is invertible if its inverse relation is also a function. For example, consider the following linear function.

$f(x)=0.5x−2 $

The graph of its inverse relation can be found by reflecting the graph of $f(x)$ across the line $y=x.$
The graph shows that the inverse relation, $g(x)=2x+4,$ is also a linear function. Therefore, $g(x)=f_{-1}(x)$ and, by definition, $f(x)=0.5x−2$ is invertible.Discussion

There is a condition that a function must satisfy in order to be invertible.

This condition implies that if a function fails the Horizontal Line Test, then it is not invertible. However, some functions do not satisfy this condition for invertibility but can become invertible by restricting their domain. For example, consider the quadratic parent function $f(x)=x_{2}.$

This function has different $x-$values where the same $y-$value is assigned. The foll how the function fails the Horizontal Line Test.

Now, the graph of $f(x)$ will be reflected across the line $y=x$ to visualize its inverse relation.

The inverse is not a function because it has input values with two different outputs assigned. The following graph further confirms this relation by showing how it fails the Vertical Line Test.

Therefore, $f(x)=x_{2}$ is not invertible because its inverse is not a function. Now, consider if the domain of $f(x)$ is restricted to $x≥0.$ In that case, every $x-$value gets assigned a unique $y-$value. With the domain of $f(x)$ defined this way, $f_{-1}(x)=x $ is a valid inverse function.

Consider a function $f(x)$ that assigns the same output to multiple inputs.

The corresponding inverse relation is not a function because it has multiple outputs assigned to the same input, which goes against the definition of function.
Therefore, to be invertible, it is a necessary condition that the function assigns a unique output to every input.

To avoid repeated output values and ensure invertibility, the domain needs to be restricted to values either

Example

Heichi is so fascinated about the invertibility of quadratic functions, that he started a study group.

He brought the following function rule.

$f(x)=21 (x−2)_{2}+8 $

Heichi asks his friends to help him identify in which intervals the function is invertible. Select all the options that correspond to these intervals. {"type":"multichoice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.5782em;vertical-align:-0.0391em;\"><\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">\u221e<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\"><<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.7719400000000001em;vertical-align:-0.13597em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.5782em;vertical-align:-0.0391em;\"><\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">\u221e<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\"><<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.7719400000000001em;vertical-align:-0.13597em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">8<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.78041em;vertical-align:-0.13597em;\"><\/span><span class=\"mord\">2<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.5782em;vertical-align:-0.0391em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\"><<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord\">\u221e<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.78041em;vertical-align:-0.13597em;\"><\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">4<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.5782em;vertical-align:-0.0391em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\"><<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord\">\u221e<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.5782em;vertical-align:-0.0391em;\"><\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">\u221e<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\"><<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.7719400000000001em;vertical-align:-0.13597em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span>"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":[0,2,4]}

It is a necessary condition for invertibility that the function assigns a unique output to every input. That being said, because of the symmetry of quadratic functions, inputs with the same distance from the vertex are assigned the same output.
*equal or less than* or *equal or greater than* the $x-$coordinate of the vertex. Since the given quadratic function is in vertex form, the vertex will be located by comparing the function with the general form.

To avoid having repeated output values, the domain needs to be restricted to values either

$Quatratic Functionf(x)=a(x−h)_{2}+kQuatratic Functionf(x)=21 (x−2)_{2}+8 Vertex(h,k)Vertex(2,8) $

Because the $x-$coordinate of the vertex is $2,$ the following intervals ensure that the function satisfies the invertibility condition.
$I.II. -∞<x≤2-I2≤x<∞ $

Three of the five given options are contained in these intervals. - $-∞<x≤-2$
- $-∞<x≤-2$
- $-I2≤x<∞$

The remaining two options are intervals where the function does not satisfy the invertibility condition. This can be verified by using the Horizontal Line Test.

Pop Quiz

The following applet shows different quadratic functions. The representation of these functions alternates between showing their graph and stating their function rule. Determine whether or not the given function is invertible in the indicated domain.

Discussion

A function can be represented by a table of values, a graph, a mapping diagram, or a function rule, among other ways. Depending on how the function is presented, finding its inverse can be done in different ways. When a function rule is given, finding the inverse algebraically is advantageous. Consider the following example function.
*expand_more*
*expand_more*
*expand_more*
*expand_more*

$f(x)=32x−1 $

There is a series of steps to follow in order to find the inverse function $f_{-1}(x).$
1

Replace $f(x)$ With $y$

To begin, since $f(x)=y$ describes the input-output relationship of the function, replace $f(x)$ with $y$ in the function rule.

$f(x)=32x−1 →y=32x−1 $

2

Switch $x$ and $y$

Because the inverse of a function reverses $x$ and $y,$ the variables can be switched. Notice that every other piece in the function rule remains the same.

$y=32x−1 switch x=32y−1 $

3

Solve for $y$

Solve the resulting equation from the previous step for $y.$ This will involve using the inverse operations.

$x=32y−1 $

▼

Solve for $y$

MultEqn

$LHS⋅3=RHS⋅3$

$3x=2y−1$

AddEqn

$LHS+1=RHS+1$

$3x+1=2y$

DivEqn

$LHS/2=RHS/2$

$23x+1 =y$

RearrangeEqn

Rearrange equation

$y=23x+1 $

4

Replace $y$ With $f_{-1}(x)$

Just as $f(x)=y$ shows the input-output relationship of $f,$ so does $f_{-1}(x)=y.$ Therefore, replacing $y$ with $f_{-1}(x)$ gives the rule for the inverse of $f.$

$y=23x+1 →f_{-1}(x)=23x+1 $

Notice that in $f,$ the input is multiplied by $2,$ decreased by $1,$ and divided by $3.$ From the rule of $f_{-1},$ it can be seen that $x$ undergoes the inverse of these operations in the reverse order. Specifically, $x$ is multiplied by $3,$ increased by $1,$ and divided by $2.$ $Vertex Formy=a(x−h)_{2}+k Switchxandyx=a(y−h)_{2}+k $

If the function rule is in vertex before switching $x$ and $y,$ the power can be isolated. Then, the operation of raising the base to the second power can be undone by taking the square root, allowing to solve for $y$ and find the inverse.
$x=a(y−h)_{2}+k$

▼

Solve for $y$

SubEqn

$LHS−k=RHS−k$

$x−k=a(y−h)_{2}$

DivEqn

$LHS/a=RHS/a$

$ax−k =(y−h)_{2}$

SqrtEqn

$LHS =RHS $

$±ax−k =y−h$

AddEqn

$LHS+h=RHS+h$

$±ax−k +h=y$

RearrangeEqn

Rearrange equation

$y=±ax−k +h$

Rewrite

Rewrite $y$ as $f_{-1}(x)$

$f_{-1}(x)=±ax−k +h$

Example

Heichi took a few members from his study group named Ignacio and Kevin to try something cool in their apartment building. In their physics class, they learned that the height of an object that falls from an initial height $h$ can be modeled using a quadratic function. They then adapted the function using the height of their building — $45$ meters.
### Answer

### Hint

### Solution

Now, the function will be restricted to the time interval where the ball starts falling until it reaches the ground. The following graph represents the resulting piece of the quadratic function after being restricted.
Finally, $y$ will be replaced with $f_{-1}(x).$
Therefore, the ball will reach Ignacio's window $1.7$ seconds after being dropped. Similarly, since Kevin's window is at a height of $20$ meters, the time that the ball takes to reach it will be found by evaluating the inverse for $x=20.$
Hence, ball will reach Kevin's window after $2.2$ seconds.

$f(x)=-5x_{2}+h⇓f(x)=-5x_{2}+45 $

Here $x$ is the time in seconds and $f(x)$ represents the height of the falling object, in meters. To test this model, Heichi goes to the top of the building and prepares to drop a tennis ball. Meanwhile, Ignacio and Kevin look attentive through their respective windows holding a chronometer as they wait for Heichi's signal. b Does the domain restriction indicated in Part A make the given function invertible? Justify the answer.

c What should be the inverse function's domain and range? Find the inverse function.

d Ignacio's window is $30$ meters above the ground, and Kevin's window is $20$ meters above the ground. Use the inverse function found in Part C to find the $x-$values corresponding to when the ball is in front of Ignacio's and Kevin's windows, respectively. Round the times to one decimal place.

a **Graph:**
**Range:** $0≤y≤45$

**Domain:** $0≤x≤3$

b **Is the Function Invertible?** Yes

**Explanation: **See solution.

c **Domain:** $0≤x≤45$

**Inverse Function:** $f_{-1}(x)=545−x $

**Range:** $0≤y≤3$

d **Time to Reach Ignacio's Window:** $1.7s$

**Time to Reach Kevin's Window:** $2.2s$

a The domain is the set of all the possible input values of a function, while the range is the set of all possible output values.

b For a function to be invertible, it must assign a unique output to every input.

c The inverse of a function reverses inputs and outputs.

d The inverse function $f_{-1}(x)$ receives as input the height of the ball with respect to the ground and returns as output the corresponding time that the ball takes to reach said height.

a First, the graph of the given quadratic function will be drawn.
The ball starts falling at $x=0$ from a height of $45$ meters. This corresponds to the vertex of the graph $(0,45).$ On the other hand, the graph intersects the $x-$axis at $(3,0).$ This means that the ball reaches the ground after $3$ seconds. Note that $-3$ is another $x-$intercept but it is discarded because it represents a negative time.

The graph shows that the domain of the restricted function is $0≤x≤3$ and the range is $0≤y≤45.$

b For a function to be invertible, it must assign a unique output value to every input. When considering the domain to be all real numbers, the function fails the Horizontal Line Test. This means that there are repeating outputs and, consequently, the function is not invertible.

However, all the $x-$values in the interval $0≤x≤3$ have unique $y-$values associated. This is why restricting the function's domain to the interval $0≤x≤3$ makes the function invertible.

c First the inverse will be found algebraically. Then, it will be explained what its domain and range must be and why. Since $f(x)=y$ describes the input-output relationship of the function, $f(x)$ will be replaced with $y$ in the function rule.

$f(x)=-5x_{2}+45→y=-5x_{2}+45 $

Now, the variables $y$ and $x$ will be switched because the inverse of a function reverses inputs and outputs. $y=-5x_{2}+45switch x=-5y_{2}+45 $

Next, the resulting equation will be solved for $y.$
$x=-5y_{2}+45$

▼

Solve for $y$

SubEqn

$LHS−45=RHS−45$

$x−45=-5y_{2}$

DivEqn

$LHS/(-5)=RHS/(-5)$

$-5x−45 =y_{2}$

MoveNegDenomToNum

Put minus sign in numerator

$5-(x−45) =y_{2}$

Distr

Distribute $-1$

$5−x+45 =y_{2}$

CommutativePropAdd

Commutative Property of Addition

$545−x =y_{2}$

SqrtEqn

$LHS =RHS $

$±545−x =y$

RearrangeEqn

Rearrange equation

$y=±545−x $

$y=±545−x ↓f_{-1}(x)=±545−x $

Recall that inverse functions are reflectionally symmetric with respect to $y=x.$ The domain of $f_{-1}(x)$ must be equal to the range of $f(x),$ which is $0≤y≤45.$ And the range $f_{-1}(x)$ must be equal to the domain of $f(x),$ which is $0≤x≤3.$ $Domain off_{-1}(x):Range off_{-1}(x): 0≤x≤450≤y≤3 $

Choosing the negative sign for the square root allows $f_{-1}(x)$ to assign negative outputs. Therefore, $f_{-1}(x)$ will only have the domain and range stated above if the positive sign is chosen. $f_{-1}(x)=545−x $

Graphing the restricted quadratic function and the found inverse together shows how their graphs are reflections of each other across the line $y=x.$ This confirms that the found inverse function is correct.
d To find the time it takes the ball to reach Igancio's window, which is at a height of $30$ meters, the inverse function will be evaluated for $x=30.$

$f_{-1}(x)=545−x $

▼

Substitute $30$ for $x$ and evaluate

Substitute

$x=30$

$f_{-1}(x)=545−30 $

SubTerm

Subtract term

$f_{-1}(x)=515 $

CalcQuot

Calculate quotient

$f_{-1}(x)=3 $

CalcRoot

Calculate root

$f_{-1}(x)=1.732050…$

RoundDec

Round to $1$ decimal place(s)

$f_{-1}(x)=1.7$

$f_{-1}(x)=545−x $

▼

Substitute $20$ for $x$ and evaluate

Substitute

$x=20$

$f_{-1}(x)=545−20 $

SubTerm

Subtract term

$f_{-1}(x)=525 $

CalcQuot

Calculate quotient

$f_{-1}(x)=5 $

CalcRoot

Calculate root

$f_{-1}(x)=2.236067…$

RoundDec

Round to $1$ decimal place(s)

$f_{-1}(x)=2.2$

Pop Quiz

The following applet shows a quadratic function and a specific point on its graph. The domain of the given function is restricted and the function is invertible. Find the coordinates of the corresponding point on the graph of its inverse function. Write the answer rounding to two decimal places.