Free Math Worksheets

- Focus on measurements taken at one fixed time. - Ask whether one height could be assigned to more than one student in the reverse relation. - An inverse function requires the original function to be one-to-one.

1. The assignment is a function when each student has exactly one recorded height at the specified measurement time. 2. The reverse assignment is not a function if two or more students have the same recorded height, because that height would correspond to more than one student. 3. The original function is invertible only if all recorded heights are distinct, making the function one-to-one.

a) Each student must have exactly one recorded height at the fixed time. b) The reverse assignment is not a function when at least two students have the same recorded height. c) All recorded heights must be different.

- Check whether any input can lead to more than one output. - Reverse the direction of the assignment and test the function definition again. - For part c, choose a school-related input for which exactly one output is defined.

1. Slot code \(\rightarrow\) price is a function because each slot code has exactly one listed price. 2. Price \(\rightarrow\) slot code is not a function because at least one price corresponds to more than one slot code. 3. One example is student \(\rightarrow\) date of birth. This is a function because each student has exactly one date of birth.

a) Yes. Each slot code has exactly one price. b) No. A price can correspond to more than one slot code. c) Example: student \(\rightarrow\) date of birth is a function.

- Move an imaginary horizontal line up and down each graph. - An invertible function can be intersected at most once by each horizontal line. - Split the noninvertible graph at the point where its direction changes. - Choose a complete increasing or decreasing branch.

1. a) Every horizontal line meets the graph of \(f\) at most once. Therefore, \(f\) is one-to-one and invertible on its entire displayed domain. 2. Some horizontal lines meet the graph of \(g\) twice. For example, \(y=0\) intersects it at \(x=-2\) and \(x=2\). Therefore, \(g\) is not invertible on its entire displayed domain. 3. b) The graph of \(g\) decreases from \(x=-3\) to its vertex at \(x=0\), then increases from \(x=0\) to \(x=3\). 4. Thus, either \([-3,0]\) or \([0,3]\) is a largest interval on which the restriction of \(g\) is invertible.

a) \(f\) is invertible on its displayed domain; \(g\) is not. b) \([-3,0]\) or \([0,3]\)

- Imagine moving a horizontal line upward across each graph. - Count how many times the line can intersect each graph. - An inverse is a function only when each output corresponds to at most one input.

1. For \(f\), a horizontal line such as \(y=1\) intersects the graph twice, at \(x=-3\) and \(x=3\). 2. Therefore, \(f\) is not one-to-one and is not invertible on its displayed domain. 3. Every horizontal line intersects the graph of \(g\) at most once. 4. Therefore, \(g\) is one-to-one and is invertible on its displayed domain.

\(f\) is not invertible because it fails the horizontal line test. \(g\) is invertible because every horizontal line intersects its graph at most once.

- Use the horizontal line test rather than the vertical line test. - Imagine moving a horizontal line through each graph. - Check whether one output can correspond to more than one input. - A strictly monotonic graph passes the test.

1. Use the horizontal line test: a function is invertible on its domain when every horizontal line intersects its graph at most once. 2. a) The graph is strictly increasing, so it passes the horizontal line test. The function is invertible. 3. b) The graph fails the test. For example, the line \(y=1\) intersects it at \(x=1\) and \(x=3\). The function is not invertible. 4. c) The square-root graph is strictly increasing on its displayed domain, so it passes the test. The function is invertible. 5. d) The parabola fails the test. For example, the line \(y=2\) intersects it twice. The function is not invertible.

a) Invertible b) Not invertible c) Invertible d) Not invertible Graphs a) and c) pass the horizontal line test; graphs b) and d) do not.

- Use the horizontal line test. - To find \(f^{-1}(2)\), ask which input produces the output \(2\). - Corresponding points on a function and its inverse have reversed coordinates.

1. a) The graph is strictly increasing for \(x\ge0\), so every horizontal line intersects it at most once. Therefore, \(f\) is invertible. 2. b) At \(x=2\), the graph has output \(2\), so \(f(2)=2\). 3. To find \(f^{-1}(2)\), find the input whose output under \(f\) is \(2\). The graph shows that this input is \(2\), so \(f^{-1}(2)=2\). 4. c) Corresponding points on inverse graphs have reversed coordinates. Therefore, \((4,5)\) on \(f\) corresponds to \((5,4)\) on \(f^{-1}\).

a) The graph is strictly increasing, so it passes the horizontal line test. b) \(f(2)=2\) and \(f^{-1}(2)=2\) c) \((5,4)\)

- An invertible function must be one-to-one. - Check whether a horizontal line can intersect a graph more than once. - Consider what happens when a graph is constant on an interval. - A graph that repeatedly rises and falls usually fails the test on a wide domain.

1. a) The graph of \(f\) is strictly decreasing on its displayed domain. Every horizontal line intersects it at most once, so \(f\) is invertible. 2. b) The graph of \(g\) rises and falls, so some horizontal lines intersect it more than once. Therefore, \(g\) is not invertible on its displayed domain. 3. c) The graph of \(h\) is constant at \(y=2\) for \(1\le x\le3\). Many inputs have the same output, so \(h\) is not one-to-one and is not invertible.

Only \(f\) is invertible. It is strictly decreasing and passes the horizontal line test. The functions \(g\) and \(h\) fail the horizontal line test.

- Function and inverse graphs are reflections across a particular line. - Check whether a point \((x,y)\) on one graph corresponds to \((y,x)\) on the other. - Use a point with easy-to-read coordinates.

1. a) The graphs are reflections of each other across the line \(y=x\). For example, \((0,1)\) on the graph of \(f\) corresponds to \((1,0)\) on the graph of \(g\). 2. Therefore, the graphs in panel a) can represent a function and its inverse. 3. b) The graphs are not reflections across \(y=x\). The point \((2,1)\) lies on \(f\), so its inverse graph would need to contain \((1,2)\). 4. The graph of \(g\) instead contains \((1,1)\), so the graphs in panel b) do not represent a function and its inverse.

a) Yes. The graphs are symmetric across \(y=x\). b) No. They are not symmetric across \(y=x\); for example, \((2,1)\) on \(f\) does not correspond to \((1,2)\) on \(g\).

- A function assigns exactly one output to each input. - An inverse function exists only when the original function is one-to-one over its domain. - Consider what happens to opposite integers when they are squared.

1. Every integer input \(x\) produces exactly one output, \(x^2\), so the relation is a function. 2. The function is not one-to-one on \(\mathbb{Z}\). For example, \(f(2)=4\) and \(f(-2)=4\). Therefore, the inverse relation would assign two inputs to the output \(4\), so it is not a function. 3. One valid restriction is \(D=\{0,1,2,3,\ldots\}\), the nonnegative integers. On this domain, each perfect square in \(W\) has exactly one input, so the restricted function is invertible.

a) Each integer input has exactly one squared output. b) No. For example, \(f(2)=f(-2)=4\), so \(f\) is not one-to-one. c) One choice is \(D=\{0,1,2,3,\ldots\}\).

- At one fixed time, can the hiker have two different distances from the cabin? - Identify ways the same distance can occur at multiple times. - A constant distance over an interval produces what kind of graph segment?

1. At any particular time, the hiker has exactly one distance from the cabin, so time \(\rightarrow\) distance is a function. 2. The reverse assignment is not a function. During the pause, one distance corresponds to many times, and during the return trip some distances are reached once while moving away and again while returning. 3. During the pause, the distance remains constant as time passes, so the graph has a horizontal segment.

a) Yes. Each time has exactly one distance. b) No. The same distance can occur at more than one time because of both the pause and the return trip. c) The graph is horizontal during the pause.

- What happens to the coordinates \((x, y)\) when a point is reflected across \(y = x\)? - How can you solve an equation for a different variable? - When does a geometric line represent the graph of a function of \(x\)? - What is special about a point that stays fixed under a reflection?

1. Reflection across \(y = x\) swaps \(x\) and \(y\). Starting with \(y = 0.4x + 3\), write \(x = 0.4y + 3\) and solve for \(y\): \(y = 2.5x - 7.5\). Thus, \(g(x) = 2.5x - 7.5\). 2. Set the functions equal: \(0.4x + 3 = 2.5x - 7.5\). Then \(10.5 = 2.1x\), so \(x = 5\), and \(y = 5\). The intersection is \((5, 5)\). The reflection swaps these two distinct lines. Because they have a unique intersection, that intersection must map to itself; fixed points of the reflection lie on \(y = x\). 3. The graph of \(k(x) = 3\) is the horizontal line \(y = 3\). Reflecting it across \(y = x\) gives the vertical line \(x = 3\), which fails the vertical line test and is not the graph of a function of \(x\).

a) \(g(x) = 2.5x - 7.5\) b) \((5, 5)\); it lies on \(y = x\), the line of reflection. c) The reflection is \(x = 3\), a vertical line, so it is not the graph of a function of \(x\).

- Interchange \(x\) and \(y\), then solve for the new output variable. - The domain of an inverse is the range of the original function. - Reflection across \(y=x\) interchanges horizontal and vertical features. - At a shared point, the two function formulas give the same output.

1. Start with \(y=\frac{2x}{x-1}\) and solve for \(x\): \(y(x-1)=2x\), so \(x(y-2)=y\) and \(x=\frac{y}{y-2}\). 2. Therefore, \(y\ne2\), so the range is \(\mathbb{R}\setminus\{2\}\), and \(f^{-1}(x)=\frac{x}{x-2}\). 3. Reflection across \(y=x\) interchanges horizontal and vertical lines. Thus, the vertical asymptote \(x=1\) becomes the horizontal asymptote \(y=1\), and the horizontal asymptote \(y=2\) becomes the vertical asymptote \(x=2\). 4. For intersections, solve \(\frac{2x}{x-1}=\frac{x}{x-2}\), with \(x\ne1,2\). This gives \(2x(x-2)=x(x-1)\), so \(x^2-3x=0\). 5. Hence, \(x=0\) or \(x=3\). The intersection points are \((0,0)\) and \((3,3)\).

1) Range: \(\mathbb{R}\setminus\{2\}\); \(f^{-1}(x)=\frac{x}{x-2}\) 2) Vertical asymptote \(x=2\); horizontal asymptote \(y=1\) 3) \((0,0)\) and \((3,3)\)

- Use the restricted domain to identify which branch of the parabola is included. - Interchange input and output, then solve for the new output. - The domain restriction determines which square-root sign to choose. - The domain and range switch when a function is inverted.

1. The domain restricts \(f\) to the left branch of an upward-opening parabola. As \(x\) increases toward \(-2\), \((x+2)^2\) decreases, so \(f\) is strictly decreasing and one-to-one. 2. Start with \(y=\frac{1}{2}(x+2)^2-4\). Then \((x+2)^2=2y+8\). 3. Because the original domain requires \(x\le-2\), choose the negative square root: \(x+2=-\sqrt{2y+8}\). Thus, \(x=-\sqrt{2y+8}-2\). 4. Interchange the variables to obtain \(f^{-1}(x)=-\sqrt{2x+8}-2\). 5. The range of \(f\) is \([-4,\infty)\), so the domain of \(f^{-1}\) is \([-4,\infty)\). The range of \(f^{-1}\) is the original domain, \((-\infty,-2]\).

1) \(f\) is strictly decreasing on \((-\infty,-2]\), so it is one-to-one. 2) \(f^{-1}(x)=-\sqrt{2x+8}-2\) 3) Domain: \([-4,\infty)\); range: \((-\infty,-2]\)

- Compare the reciprocal terms for two inputs \(a<b\). - Interchange input and output, then isolate the new output. - Use the horizontal asymptote and the behavior near \(x=-3\) to find the range. - The domain and range switch for an inverse function.

1. Suppose \(-3<a<b\). Then \(0<a+3<b+3\), so \(\frac{1}{a+3}>\frac{1}{b+3}\). It follows that \(f(a)<f(b)\). Therefore, \(f\) is strictly increasing and one-to-one. 2. Starting with \(y=5-\frac{2}{x+3}\), obtain \(5-y=\frac{2}{x+3}\), so \(x+3=\frac{2}{5-y}\) and \(x=\frac{2}{5-y}-3\). 3. Therefore, \(f^{-1}(x)=\frac{2}{5-x}-3\). 4. As \(x\) ranges over \((-3,\infty)\), \(f(x)\) takes every value less than \(5\). Thus, the inverse domain is \((-\infty,5)\), and its range is \((-3,\infty)\).

1) \(f\) is strictly increasing on \((-3,\infty)\), so it has an inverse. 2) \(f^{-1}(x)=\frac{2}{5-x}-3\) 3) Domain: \((-\infty,5)\); range: \((-3,\infty)\)

- Exponentiation reverses a natural logarithm. - The domain of an inverse is the range of the original function. - To invert the inverse, interchange input and output again. - Reflection across \(y=x\) interchanges point coordinates.

1. a) Start with \(y=\ln(x+3)\). Exponentiating gives \(e^y=x+3\), so \(x=e^y-3\). Therefore, \(f^{-1}(x)=e^x-3\). 2. The range of \(f\) is \(\mathbb{R}\), so the domain of \(f^{-1}\) is \(\mathbb{R}\). 3. b) Start with \(y=e^x-3\). Then \(y+3=e^x\), so \(x=\ln(y+3)\). Therefore, \((f^{-1})^{-1}(x)=\ln(x+3)=f(x)\). 4. c) The graph of an inverse function is the reflection of the original graph across \(y=x\). Reflecting twice across the same line returns every point to its original position, so \((f^{-1})^{-1}=f\).

a) \(f^{-1}(x)=e^x-3\), with domain \(\mathbb{R}\) b) \((f^{-1})^{-1}(x)=\ln(x+3)=f(x)\) c) The graphs are reflections across \(y=x\); two reflections return the original graph.

- Interchange input and output, then solve for the new output. - Try a line with slope \(-1\). - An inverse graph is a reflection across \(y=x\).

1. a) Start with \(y=\frac{2}{x}\). Then \(xy=2\), so \(x=\frac{2}{y}\). Interchanging the variables gives \(f^{-1}(x)=\frac{2}{x}=f(x)\). Thus, \(f\) is self-inverse. 2. b) One example is \(g(x)=-x+5\). From \(y=-x+5\), solve for \(x\): \(x=-y+5\). Therefore, \(g^{-1}(x)=-x+5=g(x)\). 3. More generally, every line \(g(x)=-x+c\) is self-inverse; the identity function \(g(x)=x\) is also self-inverse. 4. c) The graph must be symmetric about the line \(y=x\). Whenever \((a,b)\) is on the graph, \((b,a)\) must also be on the graph.

a) Yes. \(f^{-1}(x)=\frac{2}{x}=f(x)\). b) For example, \(g(x)=-x+5\), and solving for the inverse gives \(g^{-1}(x)=-x+5\). c) The graph must be symmetric about \(y=x\).

- Compare the reciprocal terms for two inputs in the domain. - Interchange input and output, then solve for the new output. - Use the horizontal asymptote and behavior near \(x=3\) to determine the range. - The domain and range switch for an inverse function.

1. Suppose \(3<a<b\). Then \(0<a-3<b-3\), so \(\frac{1}{a-3}>\frac{1}{b-3}\). Therefore, \(f(a)>f(b)\), and \(f\) is strictly decreasing and one-to-one. 2. Start with \(y=\frac{2}{x-3}+1\). Then \(y-1=\frac{2}{x-3}\), so \(x-3=\frac{2}{y-1}\) and \(x=\frac{2}{y-1}+3\). 3. Thus, \(f^{-1}(x)=\frac{2}{x-1}+3\). 4. On \((3,\infty)\), the function takes every value greater than \(1\). Therefore, the domain of \(f^{-1}\) is \((1,\infty)\), and its range is \((3,\infty)\).

\(f\) is strictly decreasing and therefore invertible. \(f^{-1}(x)=\frac{2}{x-1}+3\) Domain: \((1,\infty)\) Range: \((3,\infty)\)

- Exponential functions with positive base are strictly increasing. - A vertical shift changes the range but not monotonicity. - Use the natural logarithm to undo the exponential. - The domain and range switch for an inverse function.

1. a) The exponential function \(e^{\frac{1}{2}x}\) is strictly increasing, and a vertical shift does not change monotonicity. Therefore, \(g\) is strictly increasing and one-to-one. 2. The range of \(g\) is \((-3,\infty)\). 3. b) Start with \(y=e^{\frac{1}{2}x}-3\). Then \(y+3=e^{\frac{1}{2}x}\), so \(\ln(y+3)=\frac{1}{2}x\) and \(x=2\ln(y+3)\). 4. Therefore, \(g^{-1}(x)=2\ln(x+3)\). Its domain is \((-3,\infty)\), and its range is \(\mathbb{R}\).

a) \(g\) is strictly increasing, so it has an inverse function. b) \(g^{-1}(x)=2\ln(x+3)\), with domain \((-3,\infty)\) and range \(\mathbb{R}\)

- Inverse graphs are reflections across \(y=x\). - Try a line parallel to \(y=x\) but not equal to it. - Find the inverse by interchanging input and output. - Check whether the two line equations can be equal at any input.

1. Choose \(f(x)=x+1\). Since its slope is nonzero, it has an inverse function. 2. Solving \(y=x+1\) for \(x\) gives \(f^{-1}(x)=x-1\). 3. The two graphs are parallel lines, so they do not intersect. Geometrically, they are reflections of each other across \(y=x\). 4. Equivalently, the graph of \(f\) is parallel to \(y=x\), so it has no fixed point satisfying \(f(x)=x\). Therefore, the graphs of \(f\) and \(f^{-1}\) have no common point.

One example is \(f(x)=x+1\). Its inverse is \(f^{-1}(x)=x-1\), and the two parallel graphs are reflections across \(y=x\), so they have no common point.

- A parabola changes monotonicity at its vertex. - Restrict the domain to one complete side of the vertex. - The chosen branch determines the sign of the square root. - The domain and range switch for an inverse function.

1. a) The parabola has vertex \((-3,-1)\). It is strictly decreasing to the left of the vertex and strictly increasing to the right. 2. Therefore, the two largest possible domains are \((-\infty,-3]\) and \([-3,\infty)\). 3. b) Start with \(y=\frac{1}{3}(x+3)^2-1\). Then \((x+3)^2=3y+3\). 4. Because the chosen domain requires \(x\ge-3\), take the positive square root: \(x+3=\sqrt{3y+3}\). 5. Thus, \(x=-3+\sqrt{3y+3}\), so \(f^{-1}(x)=-3+\sqrt{3x+3}\). 6. The range of the restricted function is \([-1,\infty)\). Therefore, the inverse domain is \([-1,\infty)\), and its range is \([-3,\infty)\).

a) \((-\infty,-3]\) and \([-3,\infty)\) b) \(f^{-1}(x)=-3+\sqrt{3x+3}\), with domain \([-1,\infty)\) and range \([-3,\infty)\)

- To find an inverse, write \(y=f_a(x)\) and solve for \(x\). - Compare both the formulas and the excluded domain values. - A function and its inverse are reflections across a familiar line. - Find the vertical and horizontal asymptotes of the resulting rational function.

1. Start with \(y=\frac{ax+5}{x-3}\) and solve for \(x\): \(y(x-3)=ax+5\), so \(x(y-a)=3y+5\), and therefore \(x=\frac{3y+5}{y-a}\). 2. Thus \(f_a^{-1}(x)=\frac{3x+5}{x-a}\). This has the same rule and maximal domain as \(f_a(x)=\frac{ax+5}{x-3}\) exactly when \(a=3\). 3. A function and its inverse are reflections across \(y=x\). When they are identical, the graph is symmetric about \(y=x\). 4. For \(a=3\), the vertical asymptote is \(x=3\), and the horizontal asymptote is \(y=3\). Their intersection is \((3, 3)\), which lies on \(y=x\).

1. \(a=3\) 2. The line of symmetry is \(y=x\). The asymptotes intersect at \((3, 3)\), and this point lies on \(y=x\).

- Find the inverse of \(mx+6\) symbolically. - Compare both the slope and the y-intercept. - A line symmetric about the y-axis must be horizontal. - An inverse function requires a one-to-one original function.

1. For \(m\ne0\), start with \(y=mx+6\). Solving for \(x\) gives \(x=\frac{1}{m}y-\frac{6}{m}\), so \(g^{-1}(x)=\frac{1}{m}x-\frac{6}{m}\). 2. For \(g=g^{-1}\), matching slopes gives \(m=\frac{1}{m}\), so \(m^2=1\) and \(m=1\) or \(m=-1\). 3. If \(m=1\), matching intercepts would require \(6=-6\), which is impossible. If \(m=-1\), the inverse is \(-x+6\), the same as \(g\). Therefore, \(m=-1\). 4. A line symmetric about the y-axis must be horizontal, so it has slope \(0\). A constant function is not one-to-one and has no inverse function. Therefore, no such linear function exists.

1) \(m=-1\), so \(g(x)=-x+6\) 2) No. Y-axis symmetry requires slope \(0\), but a constant function is not invertible.

- Identify the vertex and the selected branch of the parabola. - The domain restriction determines the square-root sign. - Solve the original equation for the input variable. - The domain and range switch for an inverse function.

1. The domain restricts \(f\) to the right branch of a downward-opening parabola. Starting at the vertex \((-1,4)\), the function is strictly decreasing, so it is one-to-one and invertible. 2. Start with \(y=-\frac{1}{2}(x+1)^2+4\). Then \((x+1)^2=8-2y\). 3. Since the original domain requires \(x\ge-1\), choose the positive square root: \(x+1=\sqrt{8-2y}\). 4. Therefore, \(x=\sqrt{8-2y}-1\), so \(f^{-1}(x)=\sqrt{8-2x}-1\). 5. The range of \(f\) is \((-\infty,4]\). Thus, the inverse domain is \((-\infty,4]\), and its range is \([-1,\infty)\).

\(f\) is strictly decreasing on \([-1,\infty)\), so it is invertible. \(f^{-1}(x)=\sqrt{8-2x}-1\) Domain: \((-\infty,4]\) Range: \([-1,\infty)\)

- Use the endpoint outputs to find the range. - In an inverse table, interchange each input-output pair. - Squaring reverses the principal square root. - The inverse domain is the original range.

1. a) The square-root function is increasing. Since \(f(-4)=0\) and \(f(5)=3\), the range is \([0,3]\). 2. b) The ordered pairs for \(f\) are \((-4,0)\), \((-3,1)\), \((0,2)\), and \((5,3)\). 3. c) Interchange the coordinates to obtain the inverse pairs \((0,-4)\), \((1,-3)\), \((2,0)\), and \((3,5)\). 4. From \(y=\sqrt{x+4}\), square both sides to get \(y^2=x+4\), so \(x=y^2-4\). 5. Therefore, \(f^{-1}(x)=x^2-4\), with domain \([0,3]\).

a) Range: \([0,3]\) b) <table><tr><td>\(x\)</td><td>\(-4\)</td><td>\(-3\)</td><td>\(0\)</td><td>\(5\)</td></tr><tr><td>\(f(x)\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr></table> c) <table><tr><td>\(x\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr><tr><td>\(f^{-1}(x)\)</td><td>\(-4\)</td><td>\(-3\)</td><td>\(0\)</td><td>\(5\)</td></tr></table> \(f^{-1}(x)=x^2-4\), with domain \([0,3]\)

- Exponential functions are strictly increasing. - Divide by the leading coefficient before taking a logarithm. - The natural logarithm reverses the exponential function. - Interchange input and output after solving.

1. The exponential function is strictly increasing, and multiplying by \(4>0\) preserves that behavior. Therefore, \(f\) is strictly increasing and one-to-one. 2. Start with \(y=4e^{\frac{1}{2}x-2}\). Divide by \(4\): \(\frac{y}{4}=e^{\frac{1}{2}x-2}\). 3. Take natural logarithms: \(\ln\left(\frac{y}{4}\right)=\frac{1}{2}x-2\). 4. Solving for \(x\) gives \(x=2\ln\left(\frac{y}{4}\right)+4\). 5. Therefore, \(f^{-1}(x)=2\ln\left(\frac{x}{4}\right)+4\), with domain \((0,\infty)\).

\(f\) is strictly increasing and therefore invertible. \(f^{-1}(x)=2\ln\left(\frac{x}{4}\right)+4\)

- A logarithm requires a positive argument. - Compositions of increasing functions are increasing. - Exponentiation reverses the natural logarithm. - Interchange input and output after solving.

1. The logarithm requires \(3x+6>0\), so the domain is \((-2,\infty)\). 2. The natural logarithm is strictly increasing, and the positive linear expression \(3x+6\) is also increasing. Therefore, \(g\) is strictly increasing and one-to-one. 3. Start with \(y=\ln(3x+6)-2\). Then \(y+2=\ln(3x+6)\). 4. Exponentiating gives \(e^{y+2}=3x+6\), so \(x=\frac{1}{3}e^{y+2}-2\). 5. Therefore, \(g^{-1}(x)=\frac{1}{3}e^{x+2}-2\).

Domain of \(g\): \((-2,\infty)\) \(g\) is strictly increasing and therefore invertible. \(g^{-1}(x)=\frac{1}{3}e^{x+2}-2\)

- A positive exponential term never reaches \(0\). - Use the natural logarithm to undo the exponential. - The inverse domain is the original range. - Reflection across \(y=x\) interchanges coordinates.

1. a) Since \(e^{x-2}>0\), the function values are always greater than \(1\). As \(x\to-\infty\), the exponential term approaches \(0\), so the range is \((1,\infty)\). 2. b) Start with \(y=e^{x-2}+1\). Then \(y-1=e^{x-2}\), so \(\ln(y-1)=x-2\). 3. Therefore, \(f^{-1}(x)=\ln(x-1)+2\), with domain \((1,\infty)\). 4. c) \(f(0)=e^{-2}+1\), so the y-intercept is \((0,e^{-2}+1)\). 5. Reflection across \(y=x\) interchanges coordinates, giving the inverse point \((e^{-2}+1,0)\).

a) \((1,\infty)\) b) \(f^{-1}(x)=\ln(x-1)+2\), with domain \((1,\infty)\) c) Y-intercept: \((0,e^{-2}+1)\); inverse point: \((e^{-2}+1,0)\)

- What happens to the roles of inputs and outputs when a function is inverted? - How are the domain of a function and the range of its inverse related? - If \((a, b)\) lies on a function's graph, which ordered pair lies on the inverse function's graph?

1. An inverse function exchanges the domain and range of the original function. Therefore, \(D_g = R_f = \{x \in \mathbb{R} \mid x > 0\}\) and \(R_g = D_f = \mathbb{R}\). 2. Points on inverse functions have their coordinates reversed. Since \((0, 1)\) lies on the graph of \(f\), \((1, 0)\) lies on the graph of \(g\). Thus, \(g\) has an x-intercept at \(x = 1\).

1. \(D_g = \{x \in \mathbb{R} \mid x > 0\}\); \(R_g = \mathbb{R}\) 2. \(P'(1, 0)\); the x-intercept is \(x = 1\).

- Visualize how reflecting a graph across \(y = x\) exchanges inputs and outputs. - What happens to a horizontal line when its x- and y-coordinates are interchanged? - Relate values of the exponential function near \(0\) to inputs of the inverse function near \(0\).

1. The graph of \(g\) is the reflection of the graph of \(f\) across \(y = x\). Equivalently, the inverse of a strictly increasing one-to-one function is also strictly increasing. Therefore, \(g(x) = \log_a(x)\) is strictly increasing. 2. Reflection across \(y = x\) interchanges horizontal and vertical lines. The horizontal asymptote \(y = 0\) of \(f\) becomes the vertical asymptote \(x = 0\) of \(g\). In particular, as \(x \to 0^+\), \(g(x) \to -\infty\).

1. \(g(x) = \log_a(x)\) is strictly increasing because it is the inverse of the strictly increasing function \(f(x) = a^x\). 2. The asymptote is the y-axis, \(x = 0\).

- Substitute a point's x-coordinate and compare the output with its y-coordinate. - When the output is given, solve the function equation for the input. - Check values that make a rational function's denominator zero. - An exponential function with a positive base never has an output of \(0\).

1. For \(f\), \(f(2)=4\) and \(f(-1)=-0.5\), so \(P_1\) and \(P_4\) lie on its graph. The other listed points do not. 2. For \(g\), \(g(2)=4\), \(g(0)=-2\), and \(g(-1)=-0.5\), so \(P_1\), \(P_2\), and \(P_4\) lie on its graph. The point \(P_3\) is not in the domain because \(x=1\) makes the denominator zero. 3. For \(h\), \(h(1)=3\), so only \(P_3\) lies on its graph. 4. For \(f\): \(f(-2)=-4\), \(f(3)=13.5\), \(f(x)=0\) gives \(x=0\), and \(f(x)=27\) gives \(x=\sqrt[3]{54}\). 5. For \(g\): \(g(-2)=0\), \(g(3)=2.5\), \(g(x)=0\) gives \(x=-2\), and \(g(x)=27\) gives \(x=\frac{29}{26}\). 6. For \(h\): \(h(-2)=\frac{1}{9}\), \(h(3)=27\), \(h(x)=0\) has no real solution because \(3^x>0\), and \(h(x)=27\) gives \(x=3\).

a) For \(f\): \(P_1, P_4\) For \(g\): \(P_1, P_2, P_4\) For \(h\): \(P_3\) b) For \(f\): \(Q_1(-2, -4)\), \(Q_2(3, 13.5)\), \(Q_3(0, 0)\), \(Q_4(\sqrt[3]{54}, 27)\) For \(g\): \(Q_1(-2, 0)\), \(Q_2(3, 2.5)\), \(Q_3(-2, 0)\), \(Q_4\left(\frac{29}{26}, 27\right)\) For \(h\): \(Q_1\left(-2, \frac{1}{9}\right)\), \(Q_2(3, 27)\), no \(Q_3\), \(Q_4(3, 27)\)

- What value does the exponential part approach but never reach? - Solve the equation for the variable in the exponent. - What connection is there between strict monotonicity and invertibility?

1. Since \(e^{0.5x}>0\), \(f(x)>-4\). Also, \(f(x)\to-4\) as \(x\to-\infty\). Therefore, the range is \((-4, \infty)\), and the horizontal asymptote is \(y=-4\). 2. Start with \(y=2e^{0.5x}-4\). Then \(y+4=2e^{0.5x}\), \(\frac{y+4}{2}=e^{0.5x}\), and \(\ln\!\left(\frac{y+4}{2}\right)=0.5x\). 3. Solving for \(x\) gives \(x=2\ln\!\left(\frac{y+4}{2}\right)\). Interchanging \(x\) and \(y\), \(f^{-1}(x)=2\ln\!\left(\frac{x+4}{2}\right)\), with domain \((-4, \infty)\). 4. The function \(e^{0.5x}\) is strictly increasing because the coefficient \(0.5\) in the exponent is positive. Multiplying the outputs by \(2\) and shifting them down \(4\) units preserve this monotonicity. Thus, \(f\) is strictly increasing, so it is one-to-one and invertible.

a) Range: \((-4, \infty)\); horizontal asymptote: \(y=-4\) b) \(f^{-1}(x)=2\ln\!\left(\frac{x+4}{2}\right)\), with domain \((-4, \infty)\) c) Since \(f\) is strictly increasing on its entire domain, it is one-to-one and therefore invertible.

- Use the horizontal line test. - Split the parabola at its vertex. - The selected branch determines the square-root sign. - The inverse domain is the range of the restricted function.

1. a) The function is not one-to-one on \([0,5]\). For example, \(f(0)=1\) and \(f(4)=1\), so a horizontal line at \(y=1\) meets the graph twice. 2. b) The vertex is \((2,-1)\). The function is strictly decreasing on \([0,2]\) and strictly increasing on \([2,5]\), so these are the two largest invertibility intervals. 3. c) On the right interval, start with \(y=\frac{1}{2}(x-2)^2-1\). Then \((x-2)^2=2(y+1)\). 4. Since \(x\ge2\), take the positive square root: \(x=2+\sqrt{2(y+1)}\). 5. Therefore, \(f^{-1}(x)=2+\sqrt{2(x+1)}\). 6. On \([2,5]\), the range is \([-1,3.5]\), so the inverse domain is \([-1,3.5]\).

a) \(f\) is not one-to-one; for example, \(f(0)=f(4)=1\). b) \([0,2]\) and \([2,5]\) c) \(f^{-1}(x)=2+\sqrt{2(x+1)}\), with domain \([-1,3.5]\)

- The starting point identifies the horizontal and vertical shifts of a square-root graph. - Check the formula using other visible points. - Interchange input and output, then solve for the new output. - The inverse domain is the original range.

1. a) The starting point is \((-3,-1)\). Other integer-coordinate points include \((-2,0)\), \((1,1)\), and \((6,2)\). 2. b) A square-root graph starting at \((-3,-1)\) is shifted \(3\) units left and \(1\) unit down. Therefore, \(f(x)=\sqrt{x+3}-1\). 3. c) Start with \(y=\sqrt{x+3}-1\). Then \(y+1=\sqrt{x+3}\), so \(x=(y+1)^2-3\). 4. Therefore, \(f^{-1}(x)=(x+1)^2-3\). The range of \(f\) is \([-1,\infty)\), so the inverse domain is \([-1,\infty)\).

a) Starting point: \((-3,-1)\); for example, two other points are \((-2,0)\) and \((1,1)\) b) \(f(x)=\sqrt{x+3}-1\) c) \(f^{-1}(x)=(x+1)^2-3\), with domain \([-1,\infty)\)

- Use the horizontal line test on each graph. - Look for a horizontal line that intersects one graph more than once. - For part b, identify the turning points surrounding \(x=0\). - Keep one complete monotonic section.

1. a) A horizontal line such as \(y=0\) meets the graph of \(f\) three times, at \(x=0\) and \(x=\pm\sqrt{3}\). Therefore, \(f\) is not one-to-one on \([-3,3]\). 2. The graph of \(g\) is strictly increasing on the displayed domain, so every horizontal line meets it at most once. Therefore, \(g\) is invertible on \([-3,3]\). 3. b) The graph of \(f\) has turning points at \(x=-1\) and \(x=1\). Between these values, the graph is strictly decreasing. 4. Thus, the largest connected interval containing \(0\) on which \(f\) is invertible is \([-1,1]\).

a) \(g\) is invertible on the displayed domain; \(f\) is not. b) \([-1,1]\)

- Reflection across \(y=x\) interchanges coordinates. - Start with the endpoint of the blue graph. - The restricted domain determines the square-root sign. - The inverse domain is the original range.

1. a) Reflection across \(y=x\) interchanges point coordinates. The point \((1,-2)\) on \(f\) reflects to \((-2,1)\), and \((2,-1)\) reflects to \((-1,2)\). 2. These reflected points lie on the green curve \(g_1\), so \(g_1\) is the graph of \(f^{-1}\). 3. b) Start with \(y=(x-1)^2-2\). Then \(y+2=(x-1)^2\). 4. Since the original domain requires \(x\ge1\), take the nonnegative square root: \(x-1=\sqrt{y+2}\). 5. Thus, \(x=1+\sqrt{y+2}\), so \(f^{-1}(x)=1+\sqrt{x+2}\). 6. The range of \(f\) is \([-2,\infty)\), so the inverse domain is \([-2,\infty)\).

a) \(g_1\), because points such as \((1,-2)\) and \((2,-1)\) reflect to \((-2,1)\) and \((-1,2)\) on \(g_1\). b) \(f^{-1}(x)=1+\sqrt{x+2}\), with domain \([-2,\infty)\)

- Use the horizontal line test. - Split the graph wherever its direction changes. - Find each line equation from two endpoints. - Solve each linear equation for the input variable. - The inverse domain is the range of the corresponding piece.

1. a) A horizontal line such as \(y=2\) intersects the graph more than once, so \(f\) is not one-to-one on \([-3,5]\). 2. b) The three monotonic pieces occur on \(I_1=[-3,1]\), \(I_2=[1,3]\), and \(I_3=[3,5]\). 3. On \(I_1\), the line is \(f_1(x)=x+2\). Solving \(y=x+2\) gives \(f_1^{-1}(x)=x-2\). The range of \(f_1\) is \([-1,3]\), so the inverse domain is \([-1,3]\). 4. On \(I_2\), the line is \(f_2(x)=-x+4\). This function is self-inverse, so \(f_2^{-1}(x)=-x+4\), with domain \([1,3]\). 5. On \(I_3\), the line is \(f_3(x)=2x-5\). Solving \(y=2x-5\) gives \(f_3^{-1}(x)=\frac{1}{2}x+\frac{5}{2}\). The range of \(f_3\) is \([1,5]\), so the inverse domain is \([1,5]\).

a) A horizontal line such as \(y=2\) intersects the graph more than once. b) \(I_1=[-3,1]\), \(I_2=[1,3]\), and \(I_3=[3,5]\) c) \(I_1\): \(f_1^{-1}(x)=x-2\), domain \([-1,3]\) \(I_2\): \(f_2^{-1}(x)=-x+4\), domain \([1,3]\) \(I_3\): \(f_3^{-1}(x)=\frac{1}{2}x+\frac{5}{2}\), domain \([1,5]\)

- Use the horizontal line test. - Identify the turning points around \(x=0\). - Keep the complete monotonic section between those turning points. - An inverse function interchanges the coordinates of corresponding points.

1. a) A horizontal line such as \(y=1\) intersects the graph more than once. Therefore, different inputs produce the same output, so \(f\) is not one-to-one on \(\mathbb{R}\) and is not invertible there. 2. b) The graph has a local maximum at \(x=-1\) and a local minimum at \(x=1\). It is strictly decreasing between these turning points. 3. Therefore, the largest interval containing \(0\) on which the restriction is one-to-one is \(I=[-1,1]\). 4. c) The local maximum is the point \((-1,2)\), since \(f(-1)=2\). 5. Points on an inverse graph have their coordinates reversed, so the corresponding point on \(f^{-1}\) is \((2,-1)\).

a) The graph fails the horizontal line test, so \(f\) is not invertible on \(\mathbb{R}\). b) \(I=[-1,1]\) c) \((2,-1)\)

- Use the horizontal line test. - Solve \(y=0.5x^2+1\) for \(x\), and use the original domain to choose the correct square root. - The inverse domain is the original range. - Reflection across \(y=x\) interchanges point coordinates. - Test a point that is easy to read from the blue graph.

1. a) The graph of \(f\) is strictly increasing for \(x\ge0\), so it passes the horizontal line test and is invertible on \([0,\infty)\). 2. b) Begin with \(y=0.5x^2+1\). Then \(2(y-1)=x^2\). 3. Because the original domain requires \(x\ge0\), take the nonnegative square root: \(x=\sqrt{2y-2}\). 4. Interchanging the variables gives \(f^{-1}(x)=\sqrt{2x-2}\). 5. The range of \(f\) is \([1,\infty)\), so the domain of \(f^{-1}\) is \([1,\infty)\). 6. c) The inverse graph is the reflection of the graph of \(f\) across \(y=x\). For example, \((2,3)\) lies on \(f\), so \((3,2)\) must lie on \(f^{-1}\). The red graph \(g_1\) contains \((3,2)\), so \(g_1\) is the inverse graph.

a) The graph is strictly increasing on \([0,\infty)\), so it passes the horizontal line test. b) \(f^{-1}(x)=\sqrt{2x-2}\), with domain \([1,\infty)\) c) \(g_1\), because it is the reflection of \(f\) across \(y=x\) and contains the reflected point \((3,2)\).

- Require the expression under the square root to be nonnegative. - Use the horizontal line test. - Solve the equation for the original input, then interchange the variables. - The inverse domain is the original range. - Because this function is strictly increasing, any common point of its graph and its inverse graph lies on \(y=x\). Check solutions after squaring.

1. a) The radicand must be nonnegative, so \(x+2\ge0\), which gives \(x\ge-2\). Therefore, the domain is \([-2,\infty)\). 2. A square root is nonnegative, so the range is \([0,\infty)\). 3. b) The graph is strictly increasing on its entire domain, so it passes the horizontal line test and has an inverse. 4. c) Start with \(y=\sqrt{x+2}\). Squaring gives \(y^2=x+2\), so \(x=y^2-2\). 5. Interchanging the variables gives \(f^{-1}(x)=x^2-2\). Its domain is the range of \(f\), so the inverse domain is \([0,\infty)\). 6. d) The reflection line is \(y=x\). Because \(f\) is strictly increasing, any common point of the graphs of \(f\) and \(f^{-1}\) must lie on this line, so solve \(\sqrt{x+2}=x\). 7. Squaring gives \(x^2-x-2=0\), so \((x-2)(x+1)=0\). The candidate values are \(x=2\) and \(x=-1\). 8. The original equation requires \(x\ge0\), so \(x=-1\) is extraneous. Thus, the graphs intersect at \((2,2)\).

a) Domain: \([-2,\infty)\); range: \([0,\infty)\) b) The graph is strictly increasing, so it passes the horizontal line test. c) \(f^{-1}(x)=x^2-2\), with domain \([0,\infty)\) d) Reflection line: \(y=x\); intersection point: \((2,2)\)

- Use the horizontal line test on each panel. - Look for an output value that comes from more than one input. - A strictly increasing or strictly decreasing graph passes the test. - For a graph with separate branches, consider both branches together.

1. A function is invertible on its displayed domain exactly when it is one-to-one. Graphically, this can be checked with the horizontal line test. 2. The graph of \(f\) fails the test because some horizontal lines intersect it three times. Therefore, \(f\) is not invertible on its entire displayed domain. 3. The graph of \(g\) is strictly decreasing, so every horizontal line intersects it at most once. Therefore, \(g\) is invertible on its displayed domain. 4. The two branches of \(h\) occupy disjoint y-value ranges, and each branch is strictly decreasing. Therefore, every horizontal line intersects the graph of \(h\) at most once, so \(h\) is invertible on its displayed domain. 5. The graph of \(k\) fails the test. For example, the line \(y=1\) intersects it twice. Therefore, \(k\) is not invertible on its displayed domain.

On their displayed domains, the functions \(g\) and \(h\) are invertible because they pass the horizontal line test. The functions \(f\) and \(k\) are not invertible because each has a horizontal line that intersects its graph more than once.

- Locate the vertex of the parabola. - Use the horizontal line test. - Choose one side of the vertex where the graph is strictly monotonic. - Interchanging inputs and outputs corresponds to reflection across a particular line.

1. a) The graph fails the horizontal line test. For example, \(f(0)=f(4)=-1\), so two different inputs have the same output. 2. Therefore, \(f\) is not one-to-one on \(\mathbb{R}\) and is not invertible there. 3. b) The vertex is \((2,-3)\). To the right of the vertex, the graph is strictly increasing. 4. The largest interval of the form \([a,\infty)\) on which the restriction is one-to-one is \([2,\infty)\). 5. c) Reflect the restricted graph across the line \(y=x\).

a) The graph fails the horizontal line test; for example, \(f(0)=f(4)=-1\). b) \([2,\infty)\) c) Reflect the restricted graph across \(y=x\).

- Use the horizontal line test. - Split the parabola at its vertex. - Solve the quadratic equation for the original input. - Choose the square-root sign that matches each restricted domain. - The inverse domain is the range of the corresponding restriction.

1. a) The graph fails the horizontal line test. For example, \(f(-2)=f(4)=1.5\), so \(f\) is not one-to-one on \([-2,4]\). 2. b) The vertex occurs at \(x=1\). The graph is strictly decreasing on \(I_1=[-2,1]\) and strictly increasing on \(I_2=[1,4]\). 3. These are the two largest subintervals on which the restrictions are invertible. 4. c) Both restrictions use the formula \(f(x)=0.5(x-1)^2-3\), with their respective domains. 5. Solve \(y=0.5(x-1)^2-3\): \((x-1)^2=2y+6\), so \(x=1\pm\sqrt{2y+6}\). 6. On \(I_1\), inputs satisfy \(x\le1\), so \(f_1^{-1}(x)=1-\sqrt{2x+6}\). 7. On \(I_2\), inputs satisfy \(x\ge1\), so \(f_2^{-1}(x)=1+\sqrt{2x+6}\). 8. Each restriction has range \([-3,1.5]\), so each inverse has domain \([-3,1.5]\).

a) The graph fails the horizontal line test; for example, \(f(-2)=f(4)=1.5\). b) \(I_1=[-2,1]\) and \(I_2=[1,4]\) c) \(I_1\): \(f_1(x)=0.5(x-1)^2-3\), and \(f_1^{-1}(x)=1-\sqrt{2x+6}\), with inverse domain \([-3,1.5]\) \(I_2\): \(f_2(x)=0.5(x-1)^2-3\), and \(f_2^{-1}(x)=1+\sqrt{2x+6}\), with inverse domain \([-3,1.5]\)

- Look for an output that occurs at two different inputs. - Split the graph where it changes direction. - Find each line equation from two points. - Solve each equation for the input variable. - The inverse domain is the range of the corresponding restriction.

1. a) The horizontal line \(y=0\) intersects the graph at \(x=-3\) and \(x=1\). Therefore, \(g\) is not one-to-one on \([-4,2]\). 2. b) The graph changes direction at \(x=-1\). It is strictly decreasing on \(I_1=[-4,-1]\) and strictly increasing on \(I_2=[-1,2]\). 3. c) On \(I_1\), the line through \((-4,1)\) and \((-1,-2)\) has equation \(g_1(x)=-x-3\). 4. Solving \(y=-x-3\) for \(x\) gives \(x=-y-3\), so \(g_1^{-1}(x)=-x-3\). 5. On \(I_2\), the line through \((-1,-2)\) and \((2,1)\) has equation \(g_2(x)=x-1\). 6. Solving \(y=x-1\) for \(x\) gives \(x=y+1\), so \(g_2^{-1}(x)=x+1\). 7. Both restrictions have range \([-2,1]\), so both inverse functions have domain \([-2,1]\).

a) The graph fails the horizontal line test; for example, \(g(-3)=g(1)=0\). b) \(I_1=[-4,-1]\) and \(I_2=[-1,2]\) c) \(I_1\): \(g_1(x)=-x-3\), and \(g_1^{-1}(x)=-x-3\), with inverse domain \([-2,1]\) \(I_2\): \(g_2(x)=x-1\), and \(g_2^{-1}(x)=x+1\), with inverse domain \([-2,1]\)

- Count the inputs where the graph has output \(0\). - Split the graph at the corners where its direction changes. - First find the line equation on \([1,3]\). - Solve that equation for the input variable.

1. a) The horizontal line \(y=0\) intersects the graph at three different inputs. Therefore, \(f\) is not one-to-one on \([-3,3]\). 2. b) The graph changes direction at \(x=-1\) and \(x=1\). The three largest monotonic intervals are \([-3,-1]\), \([-1,1]\), and \([1,3]\). 3. c) On \([1,3]\), the graph is the line through \((1,-1)\) and \((3,1)\). Its slope is \(1\), so the restricted function is \(f_3(x)=x-2\). 4. Solving \(y=x-2\) for \(x\) gives \(x=y+2\). Therefore, \(f_3^{-1}(x)=x+2\). 5. The range of the restriction is \([-1,1]\), so the inverse domain is \([-1,1]\).

a) The graph fails the horizontal line test; the line \(y=0\) intersects it three times. b) \([-3,-1]\), \([-1,1]\), and \([1,3]\) c) \(f_3^{-1}(x)=x+2\), with domain \([-1,1]\)

- Look for symmetry in the graph. - Separate the two branches at the y-axis. - When solving for the original input, use the fact that the selected branch has positive inputs.

1. a) The function is symmetric about the y-axis, so \(f(x)=f(-x)\). For example, \(f(1)=f(-1)=4\). 2. Therefore, \(f\) is not one-to-one on its entire domain and is not invertible there. 3. b) The left branch is strictly increasing on \([-3,0)\), and the right branch is strictly decreasing on \((0,3]\). These are the two largest intervals on which the restrictions are invertible. 4. c) On the positive branch, start with \(y=\frac{4}{x^2}\). Then \(x^2=\frac{4}{y}\). 5. Since \(x>0\), take the positive square root: \(x=\frac{2}{\sqrt{y}}\). 6. Thus, \(f^{-1}(x)=\frac{2}{\sqrt{x}}\). The range of the positive branch is \([\frac{4}{9},\infty)\), so this is the domain of the inverse.

a) \(f\) is not invertible because \(f(x)=f(-x)\); for example, \(f(1)=f(-1)=4\). b) \([-3,0)\) and \((0,3]\) c) \(f^{-1}(x)=\frac{2}{\sqrt{x}}\), with domain \([\frac{4}{9},\infty)\)

- Locate the turning points of the graph. - Split the domain where the graph changes from increasing to decreasing or vice versa. - Reflection across \(y=x\) interchanges point coordinates.

1. a) The graph fails the horizontal line test. For example, \(f(0)=f(\sqrt{3})=f(-\sqrt{3})=0\). 2. Therefore, \(f\) is not one-to-one on \([-2.2,2.2]\) and is not invertible on the entire domain. 3. b) The graph changes direction at \(x=-1\) and \(x=1\). It is monotonic on \([-2.2,-1]\), \([-1,1]\), and \([1,2.2]\). 4. These are the three largest subintervals on which the restrictions are invertible. 5. c) Reflection across \(y=x\) interchanges coordinates, so \((1,-2)\) maps to \((-2,1)\).

a) \(f\) is not invertible because it fails the horizontal line test; for example, \(f(0)=f(\sqrt{3})=f(-\sqrt{3})=0\). b) \([-2.2,-1]\), \([-1,1]\), and \([1,2.2]\) c) \((-2,1)\)

- Choose two points with integer coordinates. - Use the slope formula and the y-intercept. - Solve the equation for the original input variable. - A function and its inverse are reflected across a standard line.

1. a) Two convenient points are \((0,-3)\) and \((2,0)\). 2. The slope is \(m=\frac{0-(-3)}{2-0}=\frac{3}{2}\), and the y-intercept is \(-3\). Therefore, \(f(x)=\frac{3}{2}x-3\). 3. b) Start with \(y=\frac{3}{2}x-3\). Then \(y+3=\frac{3}{2}x\), so \(x=\frac{2}{3}y+2\). 4. Interchanging the variables gives \(f^{-1}(x)=\frac{2}{3}x+2\). 5. c) The graphs of a function and its inverse are reflections across \(y=x\).

a) \(f(x)=\frac{3}{2}x-3\) b) \(f^{-1}(x)=\frac{2}{3}x+2\) c) \(y=x\)

- Use a graph point whose x-coordinate is \(-1\) or \(1\) to identify the base. - What operation reverses exponentiation with a fixed base? - Express both \(8\) and \(0.5\) as powers of \(2\).

1. a) The graph contains the point \((-1, 2)\). Substituting gives \(a^{-1} = 2\), so \(\frac{1}{a} = 2\) and \(a = 0.5\). Thus, \(f(x) = (0.5)^x\). 2. b) The inverse of an exponential function with base \(a\) is the logarithmic function with the same base. Therefore, \(f^{-1}(x) = \log_{0.5}(x)\). 3. c) To find \(f^{-1}(8)\), solve \((0.5)^y = 8\). Since \(0.5 = 2^{-1}\) and \(8 = 2^3\), \(2^{-y} = 2^3\). Hence, \(-y = 3\) and \(y = -3\).

a) \(a = 0.5\) b) \(f^{-1}(x) = \log_{0.5}(x)\) c) \(f^{-1}(8) = -3\)

- Points on \(y=x\) stay fixed under reflection. - Reflection across \(y=x\) interchanges point coordinates. - Solve the original equation for its input to find the inverse formula.

1. a) The graph contains the points \((1,1)\), \((2,2)\), and \((5,3)\). 2. Reflection across \(y=x\) leaves \((1,1)\) and \((2,2)\) fixed and maps \((5,3)\) to \((3,5)\). Thus, the two graphs share the points \((1,1)\) and \((2,2)\). 3. b) The given graph has formula \(f(x)=\sqrt{x-1}+1\), with domain \([1,\infty)\). 4. Solve \(y=\sqrt{x-1}+1\): \(y-1=\sqrt{x-1}\), so \((y-1)^2=x-1\) and \(x=(y-1)^2+1\). 5. Therefore, \(f^{-1}(x)=(x-1)^2+1\), with domain \([1,\infty)\).

a) The inverse graph contains \((1,1)\), \((2,2)\), and \((3,5)\). The two graphs share \((1,1)\) and \((2,2)\). b) \(f(x)=\sqrt{x-1}+1\), with domain \([1,\infty)\) \(f^{-1}(x)=(x-1)^2+1\), with domain \([1,\infty)\)

- Check whether each graph moves in only one direction. - A strictly increasing or strictly decreasing function is one-to-one. - For function \(a\), locate the point where the graph changes direction.

1. a) The graph of \(a\) decreases until \(x=1\) and then increases. It is not strictly monotonic on \([-1,3]\), so it is not invertible on that entire interval. 2. The graph of \(b\) is strictly increasing on \([-1,5]\), so \(b\) is invertible on its displayed domain. 3. The graph of \(c\) is strictly decreasing on \([0.5,5]\), so \(c\) is invertible on its displayed domain. 4. b) The vertex of the graph of \(a\) occurs at \(x=1\). The function is strictly decreasing on \([-1,1]\) and strictly increasing on \([1,3]\). 5. Therefore, either \([-1,1]\) or \([1,3]\) is a largest valid restriction interval.

a) Functions \(b\) and \(c\) are invertible on their displayed domains. Function \(a\) is not. b) One possible interval is \([1,3]\). The interval \([-1,1]\) is also valid.

- Find the minimum output on the right half of the parabola. - Use the horizontal line test. - Reflection across \(y=x\) reverses point coordinates. - Solve \(y=f(x)\) for the original input and use the restricted domain to choose the square-root sign.

1. The smallest output occurs at \(x=0\), where \(f(0)=-1\). The graph increases without bound, so the range is \([-1,\infty)\). 2. The graph is strictly increasing on \([0,\infty)\), so it passes the horizontal line test and is invertible. 3. The graph of \(f\) has y-intercept \((0,-1)\) and x-intercept \((2,0)\). Reflection across \(y=x\) reverses coordinates. 4. Therefore, the inverse graph has x-intercept \((-1,0)\) and y-intercept \((0,2)\). 5. Start with \(y=\frac{1}{4}x^2-1\). Then \(4(y+1)=x^2\). 6. Because the original domain requires \(x\ge0\), take the nonnegative square root: \(x=2\sqrt{y+1}\). 7. Thus, \(f^{-1}(x)=2\sqrt{x+1}\), with domain \([-1,\infty)\).

1. \([-1,\infty)\) 2. The graph is strictly increasing on \([0,\infty)\), so it is invertible. 3. x-intercept: \((-1,0)\); y-intercept: \((0,2)\) 4. \(f^{-1}(x)=2\sqrt{x+1}\), with domain \([-1,\infty)\)

- Recall the domain and end behavior of an exponential function. - A strictly monotonic function is one-to-one. - Use the natural logarithm to undo an exponential expression. - For \(g^{-1}(-1)\), ask which input gives \(g(x)=-1\).

1. The exponential expression is defined for every real input, so the domain is \(\mathbb{R}\). 2. Since \(e^{x-1}>0\), the graph approaches but never reaches the horizontal asymptote \(y=-2\). Therefore, the range is \((-2,\infty)\). 3. The function is strictly increasing on \(\mathbb{R}\), so it is one-to-one and has an inverse. 4. Start with \(y=e^{x-1}-2\). Then \(y+2=e^{x-1}\), so \(\ln(y+2)=x-1\). 5. Thus, \(x=\ln(y+2)+1\), and interchanging the variables gives \(g^{-1}(x)=\ln(x+2)+1\). 6. The inverse domain is the range of \(g\), so it is \((-2,\infty)\). 7. \(g^{-1}(-1)=\ln(1)+1=1\). 8. Graphically, find the input where \(g(x)=-1\). The graph contains \((1,-1)\), so \(g^{-1}(-1)=1\).

1. Domain: \(\mathbb{R}\); range: \((-2,\infty)\) 2. \(g\) is strictly increasing, so it is invertible. 3. \(g^{-1}(x)=\ln(x+2)+1\), with domain \((-2,\infty)\) 4. \(g^{-1}(-1)=1\); on the graph of \(g\), the output \(-1\) occurs at input \(1\).

- Apply the horizontal line test to each complete graph. - For separated pieces, compare their output ranges. - Check whether the output ranges of different pieces overlap. - A horizontal segment assigns the same output to many inputs.

1. a) Each line segment is strictly increasing, and the two segments have disjoint output ranges: \([-2,-1]\) and \([1,2]\). Every horizontal line intersects the complete graph at most once, so the function is invertible. 2. b) Each line segment is strictly decreasing, and the output ranges \([1,2]\) and \([-1,0]\) do not overlap. The function is invertible. 3. c) The V-shaped graph fails the horizontal line test because some horizontal lines intersect it twice. The function is not invertible. 4. d) The graph contains a horizontal segment at \(y=1\). Infinitely many inputs on that segment have the same output, so the function is not invertible.

a) Invertible b) Invertible c) Not invertible d) Not invertible

- Compare \(f(b)\) and \(f(a)\) for two inputs with \(a<b\). - A strictly increasing function cannot exchange two distinct inputs. - Interchange \(x\) and \(y\), then solve for the new output. - Reflection across \(y=x\) interchanges a point's coordinates.

1. For \(-1<a<b\), \(f(b)-f(a)=\frac{2(b-a)}{(a+1)(b+1)}>0\). Therefore, \(f\) is strictly increasing. 2. If a common point has coordinates \((u,v)\), then \(f(u)=v\) and \(f(v)=u\). If \(u<v\), strict increase would give \(v=f(u)<f(v)=u\), a contradiction; the case \(v<u\) is similar. Thus, every common point lies on \(y=x\). 3. Solve \(\frac{2x}{x+1}=x\): \(2x=x^2+x\), so \(x(x-1)=0\). The common points are \((0,0)\) and \((1,1)\), and these are the only ones. 4. To find the inverse, solve \(y=\frac{2x}{x+1}\) for \(x\): \(y(x+1)=2x\), so \(x=\frac{y}{2-y}\). Hence, \(f^{-1}(x)=\frac{x}{2-x}\). 5. Since \(f(x)=2-\frac{2}{x+1}<2\) on the domain and takes all values below \(2\), the inverse domain is \((-\infty,2)\). 6. Finally, \(f(3)=\frac{6}{4}=1.5\), so \(P\) is on the graph. Its reflection is \(Q=(1.5,3)\), and \(f^{-1}(1.5)=\frac{1.5}{0.5}=3\).

1) The common points are \((0,0)\) and \((1,1)\). 2) \(f^{-1}(x)=\frac{x}{2-x}\), with domain \((-\infty,2)\) 3) \(P=(3,1.5)\) lies on the graph of \(f\); its reflection \(Q=(1.5,3)\) lies on the graph of \(f^{-1}\).

- Use the definition of a logarithm to solve equations involving \(\log_2(x)\). - Check the domains of logarithmic, rational, and radical functions. - An equation involving \(x^2\) may have two real solutions. - When the output is given, reverse the operations to solve for the input.

1. For function (1), \(\log_2(8)=3\), so \(A\) lies on its graph. None of the other listed points do. 2. For function (2), \(\frac{2}{(-2)^2}=0.5\) and \(\frac{2}{2^2}=0.5\), so \(B\) and \(D\) lie on its graph. 3. For function (3), \(\sqrt{0+4}=2\), so \(C\) lies on its graph. 4. For \(P(1, y)\), the outputs are \(\log_2(1)=0\), \(\frac{2}{1^2}=2\), and \(\sqrt{1+4}=\sqrt{5}\). 5. For \(Q(x, 4)\), \(4=\log_2(x)\) gives \(x=16\). The equation \(4=\frac{2}{x^2}\) gives \(x^2=0.5\), so \(x=\pm\sqrt{0.5}\). The equation \(4=\sqrt{x+4}\) gives \(x=12\).

a) \(A\) lies on (1); \(B\) and \(D\) lie on (2); \(C\) lies on (3). b) For \(P(1, y)\): (1) \(y=0\), (2) \(y=2\), (3) \(y=\sqrt{5}\) For \(Q(x, 4)\): (1) \(x=16\), (2) \(x=\pm\sqrt{0.5}\), (3) \(x=12\)