Algebra 2 · F.BF.B.4 · F.BF.B.4D

Restricting a domain to get an inverse

A function sends two different inputs to the same output, so it cannot be run backwards. Can you keep part of it and run that part backwards?

You need

Finding an inverse from a ruleAlgebra 2 Graphing square root functionsAlgebra 2

Where you are 5 in Inverse functions

Functions13

Inverse functions5

One to one functions and the horizontal line test Inverse from a table or list of pairs Finding an inverse from a rule

Restricting a domain to get an inverse

The graph of an inverse

When you reach for it

Take the function

f(x) = (x - 5)^2 - 3

It sends 3 to (-2)^2 - 3 = 1 and it sends 7 to 2^2 - 3 = 1. Asked “which input gives the output 1?”, there are two honest answers, so f is not one to one and has no inverse function. The algebra of finding an inverse says the same thing in its own way: solving y = (x - 5)^2 - 3 for x produces a \pm, two inputs for one output.

Figure 1The output 1 comes from two inputs, 3 and 7, so f(x) = (x - 5)^2 - 3 has no inverse. The line x = 5 cuts the curve into two halves, and each half alone meets every horizontal line at most once.

Very often only one of the two answers is wanted. A ball passes the same height on the way up and on the way down, and the question is about the way down. A price above the best price and a price below it give the same profit, and the shop only considers raising its price. In each case you throw away the part of the domain you do not need, keep a piece on which the function is one to one, and invert that piece. The piece is a new function with the same rule and a smaller domain, and it has an inverse even though the original does not.

You reach for this move whenever a graph fails the horizontal line test, whenever solving for the input throws up a \pm, and whenever a question only makes sense on part of the domain. It is also how the square root came to exist: \displaystyle \sqrt{x} is the inverse of squaring, once squaring is cut down to the numbers that are zero or more.

The procedure

Find where the function repeats itself

Before cutting, find out how the outputs repeat. For f(x) = (x - 5)^2 - 3, step a distance d to either side of 5:

\begin{aligned} f(5 + d) &= d^2 - 3 \\ &= (-d)^2 - 3 = f(5 - d) \end{aligned}

So inputs equally far from 5 on opposite sides always share an output. Every output above -3 comes from exactly two inputs, and the output -3 comes only from x = 5, the vertex. The vertical line x = 5 is the line of symmetry of the parabola, and it is where to cut.

On the right of the cut, x \ge 5, a bigger input makes x - 5 a bigger number that is zero or more, so its square is bigger too: the function is increasing there, and an increasing function never repeats an output. On the left, x \le 5, the same argument with 5 - x shows that f is decreasing. Either half is one to one.

Choose a piece, and write the restriction into the function

The piece has to do two things. It must be one to one, or there is still no inverse. And it should still reach every output the whole function reaches, so that the inverse can answer every “which input?” question the original could raise. For a parabola that means cutting exactly at the vertex and keeping one whole side. Here the two choices are x \ge 5 and x \le 5; each keeps the vertex and reaches every output y \ge -3.

Which side to keep is a choice. The problem may name it, may name one input that must be included, or a context may decide it (only times after the peak, only prices above the best one). Whatever the choice, it is written as part of the function:

f(x) = (x - 5)^2 - 3, \quad x \ge 5

This is a different function from the unrestricted one, even though the rule is the same. Its domain is x \ge 5 and its range is y \ge -3.

Solve, and let the restriction choose the sign

Write y = (x - 5)^2 - 3 and add 3: y + 3 = (x - 5)^2. Two numbers have the square y + 3, one the opposite of the other, so

x - 5 = \pm\sqrt{y + 3}

On the whole real line the solving stops here, with two inputs. The restriction settles it. On x \ge 5 the left side x - 5 is zero or more, and \displaystyle \sqrt{y + 3} is zero or more by definition, so only the + sign can hold. Therefore \displaystyle x = 5 + \sqrt{y + 3}, and renaming the input x as usual,

f^{-1}(x) = 5 + \sqrt{x + 3}, \quad x \ge -3

The domain of the inverse is the range of the restricted f, and the range of the inverse is the restricted domain: \displaystyle 5 + \sqrt{x + 3} is never below 5. That is the quickest way to pick a sign in every problem of this kind. The outputs of the inverse must land inside the piece you kept.

Check by composing, which is where the restriction does its work

Composing one way, for x \ge -3:

f\big(f^{-1}(x)\big) = \left(\sqrt{x + 3}\right)^2 - 3 = x + 3 - 3 = x

Composing the other way needs one fact about square roots: \displaystyle \sqrt{u^2} = |u|, because the square root returns the square’s root that is zero or more (see the last part of this procedure). So

\begin{aligned} f^{-1}\big(f(x)\big) &= 5 + \sqrt{(x - 5)^2} \\ &= 5 + |x - 5| \end{aligned}

On x \ge 5 the number x - 5 is zero or more, so |x - 5| = x - 5 and the composition is x. The restriction is exactly what makes the check work. Without it, f^{-1}\big(f(1)\big) = f^{-1}(13) = 5 + 4 = 9, not 1: the inverse undoes the restricted function only.

With a number: \displaystyle f^{-1}(1) = 5 + \sqrt{4} = 7, and indeed f(7) = 1. Of the two inputs 3 and 7, the right half gives back 7, the one that lies in it.

Figure 2The right half, f(x) = (x - 5)^2 - 3 for x \ge 5, and its inverse \displaystyle f^{-1}(x) = 5 + \sqrt{x + 3} for x \ge -3. The inverse gives back 7, not 3, for the output 1.

The square root is a restricted inverse

The plainest example of all is squaring, s(x) = x^2, which is not one to one: s(-3) = s(3) = 9. Restrict it to x \ge 0. On that piece squaring is increasing, its range is every number that is zero or more, and its inverse sends each y \ge 0 to the one number x \ge 0 whose square is y.

That inverse is the square root. The symbol \displaystyle \sqrt{x} is defined as the inverse of squaring restricted to x \ge 0, so

\begin{aligned} s(x) &= x^2, \quad x \ge 0 \\ s^{-1}(x) &= \sqrt{x}, \quad x \ge 0 \end{aligned}

This is why \displaystyle \sqrt{9} = 3 and not -3. It is not an arbitrary habit: \displaystyle \sqrt{\phantom{x}} belongs to the piece x \ge 0, so its outputs are zero or more. Had the piece x \le 0 been kept instead, the inverse would be \displaystyle -\sqrt{x}, and \displaystyle -\sqrt{9} = -3. Solving x^2 = 9 on the whole real line needs both pieces at once, which is why its answer is written \displaystyle x = \pm\sqrt{9}, and why the shape of the square root graph is half of a parabola lying on its side.

The same choice explains \displaystyle \sqrt{u^2} = |u|. The square root always lands in the piece x \ge 0, so for a negative u it returns -u, not u: \displaystyle \sqrt{(-4)^2} = \sqrt{16} = 4. Every even power works like squaring (the fourth root is the inverse of x^4 cut down to x \ge 0; see Higher roots), while odd powers such as x^3 are one to one already, so the cube root needs no restriction and accepts negative numbers.

Worked examples

Example 1

The other half of the same parabola

Let g(x) = (x - 5)^2 - 3, x \le 5: the same rule as f above, cut on the other side. Find g^{-1} with its domain and range, check it, and find g^{-1}(1).

Solution

The left half is one to one (it is decreasing, from the procedure) and reaches every output y \ge -3. Solving starts as before: y + 3 = (x - 5)^2, so \displaystyle x - 5 = \pm\sqrt{y + 3}. Now the restriction is x \le 5, so x - 5 is zero or less, and only the - sign can hold: \displaystyle x = 5 - \sqrt{y + 3}.

g^{-1}(x) = 5 - \sqrt{x + 3}, \quad x \ge -3

Its range is y \le 5, since 5 minus a number that is zero or more is at most 5, and that is the domain of g as it must be.

Check: for x \ge -3, \displaystyle g\big(g^{-1}(x)\big) = \left(-\sqrt{x + 3}\right)^2 - 3, which is x. For x \le 5, g^{-1}\big(g(x)\big) = 5 - |x - 5|, which is 5 - (5 - x) = x because x - 5 \le 0 means |x - 5| = 5 - x.

Then \displaystyle g^{-1}(1) = 5 - \sqrt{4} = 3, and indeed g(3) = 4 - 3 = 1. The right half gave back 7 and the left half gives back 3. Neither is wrong: they are inverses of two different functions.

Figure 3The left half, g(x) = (x - 5)^2 - 3 for x \le 5, and its inverse \displaystyle g^{-1}(x) = 5 - \sqrt{x + 3} for x \ge -3. The same rule, cut on the other side, gives back 3 for the output 1.

Example 2

A parabola that opens downward

Find the inverse of \displaystyle p(x) = 6 - \frac{1}{2}x^2, x \ge 0, with its domain and range, and find p^{-1}(4).

Solution

The vertex is (0, 6) and the parabola opens downward. As x grows from 0, x^2 grows and p falls, so the piece is one to one. Its outputs start at p(0) = 6 and fall without end: the range is y \le 6.

Write \displaystyle y = 6 - \frac{1}{2}x^2. Then \displaystyle \frac{1}{2}x^2 = 6 - y, so x^2 = 12 - 2y and \displaystyle x = \pm\sqrt{12 - 2y}. The piece has x \ge 0, so the + sign:

p^{-1}(x) = \sqrt{12 - 2x}, \quad x \le 6

The domain x \le 6 is the range of p, and it is also exactly where 12 - 2x \ge 0. This time the domain of the inverse runs downward, because the parabola opens downward. The range of p^{-1} is y \ge 0, the piece that was kept.

\displaystyle p^{-1}(4) = \sqrt{4} = 2, and p(2) = 6 - 2 = 4. Check: \displaystyle p\big(p^{-1}(x)\big) = 6 - \frac{1}{2}(12 - 2x) = x and \displaystyle p^{-1}\big(p(x)\big) = \sqrt{12 - 12 + x^2} = \sqrt{x^2} = |x| = x because x \ge 0.

Example 3

Choosing the piece when the rule is not in vertex form

Restrict k(x) = x^2 + 6x + 7 to the largest domain that contains x = 0 and on which k is one to one. Then find the inverse.

Solution

The cut goes at the vertex, so first find it by completing the square: half of 6 is 3, and 3^2 = 9, so

x^2 + 6x + 7 = (x^2 + 6x + 9) - 2 = (x + 3)^2 - 2

The vertex is at x = -3. The two largest one to one pieces are x \le -3 and x \ge -3, and 0 lies in the second. So the restricted function is k(x) = (x + 3)^2 - 2, x \ge -3, with range y \ge -2.

Solve: y + 2 = (x + 3)^2, so \displaystyle x + 3 = \pm\sqrt{y + 2}. On the piece x + 3 \ge 0, so the + sign:

k^{-1}(x) = -3 + \sqrt{x + 2}, \quad x \ge -2

Check: k(0) = 7 and k^{-1}(7) = -3 + 3 = 0. Also k(-1) = 1 - 6 + 7 = 2 and k^{-1}(2) = -3 + 2 = -1. The input -5 gives the same output, k(-5) = 25 - 30 + 7 = 2, but -5 is not in the piece, and the inverse rightly does not return it.

Example 4

A number in front, and the left branch

Find the inverse of m(x) = 2x^2 - 12x + 11, x \le 3.

Solution

Take the 2 out of the two x terms, then complete the square inside the bracket. Adding 9 inside the bracket adds 2 \cdot 9 = 18 to the whole, so 18 is taken away outside:

\begin{aligned} 2x^2 - 12x + 11 &= 2(x^2 - 6x) + 11 \\ &= 2(x^2 - 6x + 9) - 18 + 11 \\ &= 2(x - 3)^2 - 7 \end{aligned}

The vertex is (3, -7) and the parabola opens upward, so x \le 3 is the whole left half: one to one, with range y \ge -7.

Solve y = 2(x - 3)^2 - 7: \displaystyle (x - 3)^2 = \frac{y + 7}{2}, so \displaystyle x - 3 = \pm\sqrt{\frac{y + 7}{2}}. On the piece x - 3 \le 0, so the - sign:

m^{-1}(x) = 3 - \sqrt{\frac{x + 7}{2}}, \quad x \ge -7

Check: m(1) = 2 - 12 + 11 = 1 and \displaystyle m^{-1}(1) = 3 - \sqrt{4} = 1. Also m(-1) = 2 + 12 + 11 = 25 and \displaystyle m^{-1}(25) = 3 - \sqrt{16} = -1. In general \displaystyle m^{-1}\big(m(x)\big) = 3 - \sqrt{(x - 3)^2} = 3 - |x - 3| = 3 - (3 - x) = x for x \le 3.

Example 5

An absolute value, cut at its corner

The function v(x) = 4 - 3\lvert x + 2 \rvert is not one to one. Cut it into two one to one pieces and find the inverse of each.

Solution

The absolute value \lvert x + 2 \rvert is the distance from x to -2, so the inputs -2 + d and -2 - d always share the output 4 - 3d. For example v(0) = v(-4) = -2. The graph is a V turned upside down with its corner at (-2, 4), and the cut goes at the corner.

On the right piece, call it v_1(x) = v(x) for x \ge -2, the number x + 2 is zero or more, so \lvert x + 2 \rvert = x + 2 and v_1(x) = 4 - 3(x + 2) = -3x - 2. Its outputs start at 4 and fall, so its range is y \le 4. Solving y = -3x - 2 gives \displaystyle x = -\frac{y + 2}{3}:

v_1^{-1}(x) = -\frac{x + 2}{3}, \quad x \le 4

On the left piece, v_2(x) = v(x) for x \le -2, the number x + 2 is zero or less, so \lvert x + 2 \rvert = -(x + 2) and v_2(x) = 4 + 3(x + 2) = 3x + 10, again with range y \le 4. Solving y = 3x + 10 gives

v_2^{-1}(x) = \frac{x - 10}{3}, \quad x \le 4

For the output -2: v_1^{-1}(-2) = 0 and \displaystyle v_2^{-1}(-2) = \frac{-12}{3} = -4, the two inputs, one from each piece. No \pm appeared this time. The restriction did its work earlier, by deciding how to remove the absolute value bars before any solving began.

Example 6

A ball on the way down

A ball is thrown upward from a ledge. Its height after t seconds is h(t) = 20 + 15t - 5t^2 meters, from the throw until it lands. (a) Find the domain of h and explain why it has no inverse. (b) Write the time as a function of the height on the way down, and on the way up. (c) When is the ball at 26.25 meters? When is it at 11.25 meters?

Solution

(a) Complete the square, taking out -5:

\begin{aligned} 20 + 15t - 5t^2 &= -5(t^2 - 3t) + 20 \\ &= -5(t - 1.5)^2 + 11.25 + 20 \\ &= 31.25 - 5(t - 1.5)^2 \end{aligned}

The ball peaks at 31.25 meters after 1.5 seconds. It lands when h = 0: (t - 1.5)^2 = 6.25, so t - 1.5 = \pm 2.5, and t = 4 (the value t = -1 is before the throw). So the domain is 0 \le t \le 4. Heights between 20 and 31.25 meters are passed twice, once rising and once falling, so h is not one to one.

(b) Cut at the peak. The rising piece is 0 \le t \le 1.5, with heights 20 \le h \le 31.25; the falling piece is 1.5 \le t \le 4, with heights 0 \le h \le 31.25. Solving for t, keeping the letters because they carry units: \displaystyle (t - 1.5)^2 = \frac{31.25 - h}{5}, so \displaystyle t = 1.5 \pm \sqrt{\frac{31.25 - h}{5}}. On the way down t \ge 1.5, so

\begin{aligned} t &= 1.5 + \sqrt{\frac{31.25 - h}{5}}, \\ & \qquad 0 \le h \le 31.25 \end{aligned}

and on the way up t \le 1.5, so \displaystyle t = 1.5 - \sqrt{\frac{31.25 - h}{5}}, for 20 \le h \le 31.25. Each takes a height in meters and gives a time in seconds.

(c) For 26.25 meters, \displaystyle \sqrt{\frac{5}{5}} = 1: the ball is there at t = 0.5 seconds on the way up and t = 2.5 seconds on the way down. Check: h(0.5) = 20 + 7.5 - 1.25 = 26.25 and h(2.5) = 20 + 37.5 - 31.25 = 26.25.

For 11.25 meters, \displaystyle \sqrt{\frac{20}{5}} = 2, and the way down gives t = 3.5 seconds. The rising formula would give 1.5 - 2 = -0.5, but 11.25 is not in its domain: the ball starts at 20 meters, so it is below that only on the way down. Check: h(3.5) = 20 + 52.5 - 61.25 = 11.25.

Example 7

A reciprocal squared

Let r(x) = \dfrac{1}{(x - 2)^2}. Show that r is not one to one, restrict it to x > 2, and find the inverse of the restricted function.

Solution

\displaystyle r(2 + d) = \frac{1}{d^2} = r(2 - d), so inputs equally far from 2 share an output; for example r(1) = r(3) = 1. The function has no value at x = 2, and that is where it is cut. On x > 2, a bigger input makes (x - 2)^2 bigger and r smaller, so the piece is one to one. Its outputs are every positive number: huge just to the right of 2, close to 0 far away, never 0 itself. The range is y > 0.

Solve \displaystyle y = \frac{1}{(x - 2)^2}. Since y \ne 0, take reciprocals: \displaystyle (x - 2)^2 = \frac{1}{y}, so \displaystyle x - 2 = \pm\frac{1}{\sqrt{y}}. On the piece x - 2 > 0, so the + sign:

r^{-1}(x) = 2 + \frac{1}{\sqrt{x}}, \quad x > 0

Both inequalities are strict: the piece leaves out x = 2, and the inverse never outputs 2.

Check: r(3) = 1 and r^{-1}(1) = 3; \displaystyle r(2.5) = \frac{1}{0.25} = 4 and \displaystyle r^{-1}(4) = 2 + \frac{1}{2} = 2.5. In general \displaystyle r\big(r^{-1}(x)\big) = \frac{1}{\left(1/\sqrt{x}\right)^2} = x and \displaystyle r^{-1}\big(r(x)\big) = 2 + \sqrt{(x - 2)^2} = 2 + |x - 2| = x for x > 2.

Boundaries

The piece must be one to one, and for a curve that turns, it stops at the turn. A piece of f(x) = (x - 5)^2 - 3 that reaches across x = 5, such as 3 \le x \le 8, still contains 3 and 7, which share the output 1. It has no inverse.

The inverse belongs to the restricted function, not the original. f^{-1}\big(f(x)\big) = x holds only for inputs in the piece that was kept. For f restricted to x \ge 5, f^{-1}\big(f(1)\big) = 9.

The choice of piece is not unique, and different pieces give different inverses. The right and left halves of the same parabola have the inverses \displaystyle 5 + \sqrt{x + 3} and \displaystyle 5 - \sqrt{x + 3}. “The inverse of x^2” only makes sense once a piece is named; the square root is the name for the piece x \ge 0.

A piece smaller than needed loses outputs. Restricting f to x \ge 7 gives a one to one function, but its range is only y \ge 1, so its inverse cannot answer for any output from -3 up to 1.

Some functions have no useful piece. A constant function gives the same output everywhere, so no piece with two or more inputs is one to one.

A repeating function has infinitely many pieces. The sine and cosine repeat forever, so a standard slice has to be agreed on before they can be inverted.

A one to one piece always has an inverse, but not always an easy formula. The function x^3 - 3x is one to one on x \ge 1, yet solving y = x^3 - 3x for x is a full cubic equation, far beyond taking one square root. The inverse exists and can be read from a graph or a table.

Where this goes next

Every figure in this lesson shows a piece and its inverse as mirror images in the line y = x; why that happens, and how to draw an inverse straight from a graph, is the graph of an inverse. The inverses found here are radical functions, so their graphs are the half parabolas of square root graphs. The same move made on the repeating functions of trigonometry, keeping one agreed slice of the sine, cosine or tangent, gives the inverse trigonometric functions, which turn a known ratio back into an angle. All of this sits inside the wider topic of inverse functions.

Exercises

The restriction is given

Find the inverse of each restricted function, with its domain.

f(x) = (x + 3)^2 + 2, x \ge -3
g(x) = (x - 1)^2 - 4, x \le 1
h(x) = x^2 + 7, x \ge 0
k(x) = 12 - x^2, x \le 0
p(x) = 3(x - 2)^2, x \ge 2
\displaystyle q(x) = \frac{1}{2}(x + 4)^2 - 1, x \le -4

Choosing the piece

Restrict f(x) = x^2 - 8x + 9 to the largest domain containing x = 6 on which it is one to one, and find the inverse.
Restrict g(x) = x^2 + 6x to the largest domain containing x = -5 on which it is one to one, and find the inverse.
Restrict h(x) = 2x^2 - 4x + 5 to the largest domain containing x = 0 on which it is one to one, and find the inverse.
Restrict k(x) = 5 - 2x - x^2 to the largest domain containing x = 1 on which it is one to one, and find the inverse.
Cut v(x) = 2\lvert x - 4 \rvert + 3 into two one to one pieces and find the inverse of each.
Find the inverse of w(x) = \dfrac{9}{x^2}, x < 0.

Evaluating and checking

For f(x) = (x - 5)^2 - 3, find both inputs whose output is 13. Then find the value at 13 of the inverse of the right half (x \ge 5) and of the inverse of the left half (x \le 5).
Show by composing both ways that g(x) = (x + 2)^2, x \le -2, and \displaystyle u(x) = -2 - \sqrt{x} are inverses.
Squaring restricted to x \le 0, s(x) = x^2, has the inverse \displaystyle s^{-1}(x) = -\sqrt{x}. Find s^{-1}(49) and \displaystyle s^{-1}\!\left(\frac{9}{16}\right), and check each.

In context

A stone is dropped from a bridge 45 meters above a river, and its height after t seconds is h(t) = 45 - 5t^2 meters, for 0 \le t \le 3. Write the time as a function of the height, with its domain, and find when the stone is 25 meters above the river.
A shop’s monthly profit at a price of x dollars is P(x) = 1800 - 2(x - 30)^2 dollars, for 0 \le x \le 60. (a) Explain why the price cannot be found from the profit alone. (b) Restrict to 30 \le x \le 60 and write the price as a function of the profit. (c) On that piece, which price gives a profit of 1000 dollars, and which other price gives the same profit?

Reasoning

Use f(h + d) = f(h - d) to explain why a one to one piece of the parabola f(x) = a(x - h)^2 + k that is an interval cannot contain numbers on both sides of h.
Mateo restricts f(x) = (x - 5)^2 - 3 to x \ge 7. Is his function one to one? Find its inverse with its domain, and say what has been lost compared with the restriction x \ge 5.
Explain, using the definition of \displaystyle \sqrt{x} as a restricted inverse, why \displaystyle \sqrt{x^2} = |x| for every real x, and why \displaystyle \sqrt{x^2} = x is false for some x.
Show that s(x) = x^2, restricted to the inputs x \le -2 together with 0 \le x \le 1, is one to one, and describe its inverse and the domain of that inverse.
Show that f(x) = a(x - h)^2 + k with a > 0, restricted to x \ge h, has the inverse \displaystyle f^{-1}(x) = h + \sqrt{\frac{x - k}{a}}, x \ge k. What changes when the restriction is x \le h, and what changes when a < 0?

Answers

\displaystyle f^{-1}(x) = -3 + \sqrt{x - 2}, x \ge 2. From y - 2 = (x + 3)^2, \displaystyle x + 3 = \pm\sqrt{y - 2}, and x + 3 \ge 0 on the piece keeps the + sign. The range of f is y \ge 2.
\displaystyle g^{-1}(x) = 1 - \sqrt{x + 4}, x \ge -4. From y + 4 = (x - 1)^2, and x - 1 \le 0 on the piece keeps the - sign.
\displaystyle h^{-1}(x) = \sqrt{x - 7}, x \ge 7. From x^2 = y - 7 with x \ge 0.
\displaystyle k^{-1}(x) = -\sqrt{12 - x}, x \le 12. From x^2 = 12 - y with x \le 0, so the - sign. The range of k is y \le 12. Check: k(-2) = 8 and \displaystyle -\sqrt{4} = -2.
\displaystyle p^{-1}(x) = 2 + \sqrt{\frac{x}{3}}, x \ge 0. From \displaystyle (x - 2)^2 = \frac{y}{3} with x - 2 \ge 0.
\displaystyle q^{-1}(x) = -4 - \sqrt{2x + 2}, x \ge -1. From \displaystyle y + 1 = \frac{1}{2}(x + 4)^2, (x + 4)^2 = 2y + 2, and x + 4 \le 0 keeps the - sign. Check: q(-6) = 2 - 1 = 1 and \displaystyle -4 - \sqrt{4} = -6.
f(x) = (x - 4)^2 - 7, so the piece is x \ge 4 and \displaystyle f^{-1}(x) = 4 + \sqrt{x + 7}, x \ge -7. Check: f(6) = 36 - 48 + 9 = -3 and \displaystyle 4 + \sqrt{4} = 6.
g(x) = (x + 3)^2 - 9, so the piece is x \le -3 and \displaystyle g^{-1}(x) = -3 - \sqrt{x + 9}, x \ge -9. Check: g(-5) = 25 - 30 = -5 and \displaystyle -3 - \sqrt{4} = -5.
h(x) = 2(x - 1)^2 + 3, so the piece is x \le 1 and \displaystyle h^{-1}(x) = 1 - \sqrt{\frac{x - 3}{2}}, x \ge 3. Check: h(0) = 5 and \displaystyle 1 - \sqrt{1} = 0; h(-1) = 11 and \displaystyle 1 - \sqrt{4} = -1.
k(x) = 6 - (x + 1)^2, so the piece is x \ge -1 and \displaystyle k^{-1}(x) = -1 + \sqrt{6 - x}, x \le 6. The parabola opens downward, so the range of the piece is y \le 6. Check: k(1) = 2 and \displaystyle -1 + \sqrt{4} = 1.
Cut at the corner x = 4. For x \ge 4, v(x) = 2x - 5 with inverse \displaystyle \frac{x + 5}{2}, x \ge 3. For x \le 4, v(x) = 11 - 2x with inverse \displaystyle \frac{11 - x}{2}, x \ge 3. Check: v(6) = 7 and \displaystyle \frac{12}{2} = 6; v(1) = 9 and \displaystyle \frac{2}{2} = 1.
\displaystyle w^{-1}(x) = -\frac{3}{\sqrt{x}}, x > 0. From \displaystyle x^2 = \frac{9}{y}, \displaystyle x = \pm\frac{3}{\sqrt{y}}, and x < 0 keeps the - sign. The range of w on x < 0 is every positive number. Check: w(-3) = 1 and \displaystyle -\frac{3}{1} = -3.
(x - 5)^2 = 16, so x = 9 or x = 1. The right-half inverse gives \displaystyle 5 + \sqrt{16} = 9 and the left-half inverse gives \displaystyle 5 - \sqrt{16} = 1: each returns the input in its own piece.
For x \ge 0, \displaystyle g(u(x)) = (-\sqrt{x})^2 = x. For x \le -2, \displaystyle u(g(x)) = -2 - \sqrt{(x + 2)^2} = -2 - |x + 2| = -2 + (x + 2) = x since x + 2 \le 0. Both give x, so they are inverses.
s^{-1}(49) = -7, and (-7)^2 = 49 with -7 \le 0. \displaystyle s^{-1}\!\left(\frac{9}{16}\right) = -\frac{3}{4}, and \displaystyle \left(-\frac{3}{4}\right)^2 = \frac{9}{16}.
\displaystyle t = \sqrt{\frac{45 - h}{5}}, for 0 \le h \le 45. From 5t^2 = 45 - h with t \ge 0. At 25 meters, \displaystyle t = \sqrt{4} = 2 seconds. Check: 45 - 20 = 25.
(a) Prices equally far from 30 give the same profit, for example P(10) = P(50), so a profit does not name one price. (b) \displaystyle x = 30 + \sqrt{\frac{1800 - P}{2}}, for 0 \le P \le 1800, from \displaystyle (x - 30)^2 = \frac{1800 - P}{2} with x - 30 \ge 0. (c) \displaystyle \sqrt{400} = 20, so the price is 50 dollars; the other price is 30 - 20 = 10 dollars. Check: 1800 - 2(400) = 1000.
Suppose the piece contains p < h and q > h. Let d be the smaller of h - p and q - h, so d > 0. Being an interval, the piece contains every number from p to q, so it contains both h - d and h + d. These are different inputs with f(h - d) = ad^2 + k = f(h + d), so the piece is not one to one.
Yes: x \ge 7 lies inside the right half, where f is increasing. Its range is y \ge f(7) = 1, so the inverse is \displaystyle 5 + \sqrt{x + 3} with domain only x \ge 1. The outputs from -3 up to 1, produced by the inputs from 5 up to 7, now have no input to return to.
\displaystyle \sqrt{x} is the inverse of squaring on the piece x \ge 0, so \displaystyle \sqrt{x^2} is the number in that piece whose square is x^2. Both x and -x have the square x^2, and the one that is zero or more is |x|. When x < 0, x itself is not in the piece: \displaystyle \sqrt{(-5)^2} = 5 \ne -5.
Inputs x \le -2 give outputs y \ge 4, and there squaring is decreasing, so one to one. Inputs 0 \le x \le 1 give outputs 0 \le y \le 1, increasing, so one to one. The two output sets do not overlap, so no output comes from both parts. The inverse is \displaystyle \sqrt{x} for 0 \le x \le 1 and \displaystyle -\sqrt{x} for x \ge 4; its domain is 0 \le x \le 1 together with x \ge 4, and outputs between 1 and 4 are lost.
From y = a(x - h)^2 + k, \displaystyle (x - h)^2 = \frac{y - k}{a}, which is zero or more on the range y \ge k, so \displaystyle x - h = \pm\sqrt{\frac{y - k}{a}}, and x - h \ge 0 keeps the + sign. On x \le h the sign is -: \displaystyle h - \sqrt{\frac{x - k}{a}}, with the same domain x \ge k. When a < 0 the parabola opens downward, its range is y \le k, and the domain of the inverse becomes x \le k, where \displaystyle \frac{x - k}{a} \ge 0; the sign is still chosen by the side that was kept.