Free math worksheets

- Rewrite the composition notation using nested function notation. - The inner function acts on the original input first.

1. Composition notation means \((f\circ g)(4)=f(g(4))\). 2. Therefore, \(g\) is applied first.

\(g\) is applied first.

- Start with the inner function value that is already given. - Feed that output into the outer function.

1. The inner value is given directly: \(f(3)=8\). 2. Use that output as the input of \(g\): \(g(8)=1\). 3. Therefore, \(g(f(3))=1\).

\(1\)

- Substitute the entire inner function into the outer function. - Check both the domain of the inner function and the input restrictions of the outer function. - Recall the domains of exponential and square-root functions.

1. For \(f=u\circ v\), \(f(x)=u(v(x))=\sqrt{e^x+2}\). Since \(e^x>0\) for every real \(x\), the radicand is always positive. Therefore, \(D_f=\mathbb{R}\). 2. For \(g=v\circ u\), \(g(x)=v(u(x))=e^{\sqrt{x+2}}\). The square root requires \(x+2\ge0\), so \(D_g=[-2, \infty)\).

\(f(x)=\sqrt{e^x+2}\), \(D_f=\mathbb{R}\) \(g(x)=e^{\sqrt{x+2}}\), \(D_g=[-2, \infty)\)

- Evaluate the inner function first. - Use the result as the input of the outer function. - The notation \(f\circ g\) means \(f(g(x))\).

1. \(g(3)=-1\), so \(f(g(3))=f(-1)=3\cdot(-1)^2-4=-1\). 2. \(f(-2)=3\cdot(-2)^2-4=8\), so \(g(f(-2))=g(8)=-6\). 3. \(g(1)=1\), so \((f\circ g)(1)=f(1)=-1\). 4. \(g(5)=-3\), so \((g\circ g)(5)=g(-3)=5\).

a) \(-1\) b) \(-6\) c) \(-1\) d) \(5\)

- Evaluate the inner function first. - Recall that \(e^0=1\). - Pay attention to the order in which the functions are applied.

1. \(q(0)=1\), so \(p(q(0))=p(1)=e^0=1\). 2. \(p(1)=1\), so \(q(p(1))=q(1)=5\). 3. \(q(0.25)=2\), so \((p\circ q)(0.25)=p(2)=e\). 4. \(q(-1)=-3\), so \(q(q(-1))=q(-3)=-11\).

a) \(1\) b) \(5\) c) \(e\) d) \(-11\)

- Identify the inner and outer function in each composition. - One input where the outputs differ is enough to prove that two functions are not identical.

1. Composing in the two orders gives \(u(x)=f(g(x))=e^{x^2+2}\) and \(v(x)=g(f(x))=(e^x)^2+2=e^{2x}+2\). 2. At \(x=0\), \(u(0)=e^2\) and \(v(0)=e^0+2=3\). 3. Since \(e^2\ne3\), the functions are not equal. Therefore, \(f\circ g\ne g\circ f\).

a) \(u(x)=e^{x^2+2}\) and \(v(x)=e^{2x}+2\) b) \(u(0)=e^2\) and \(v(0)=3\), so \(f\circ g\ne g\circ f\).

- In a composition, work from right to left. - Replace the input of each outer function with the entire inner expression. - Trigonometric inputs are measured in radians unless stated otherwise.

1. Work from the inside out: \(f(x)=u(v(w(x)))=u(\cos^2(x))=3\cos^2(x)-2\). 2. Similarly, \(g(x)=w(v(u(x)))=w((3x-2)^2)=\cos((3x-2)^2)\). 3. Since \(\cos(\pi)=-1\), \(f(\pi)=3(-1)^2-2=1\).

a) \(f(x)=3\cos^2(x)-2\) and \(g(x)=\cos((3x-2)^2)\) b) \(f(\pi)=1\)

- Identify the operation performed first; it belongs to the inner function. - Replace the inner expression with a temporary variable to see the outer function. - More than one decomposition can be correct.

1. For part a, choose \(v(x)=3x^2-5\) and \(u(x)=e^x\). 2. For part b, choose \(v(x)=\sin(x)-4\) and \(u(x)=\frac{2}{x}\). 3. For part c, choose \(v(x)=\frac{2x+1}{x^2}\) and \(u(x)=\ln(x)\). 4. For part d, choose \(v(x)=\sqrt{x}+1\) and \(u(x)=4x^5\). 5. Substituting each inner function into its outer function reproduces the original expression. Other valid decompositions may also be possible.

a) \(v(x)=3x^2-5\), \(u(x)=e^x\) b) \(v(x)=\sin(x)-4\), \(u(x)=\frac{2}{x}\) c) \(v(x)=\frac{2x+1}{x^2}\), \(u(x)=\ln(x)\) d) \(v(x)=\sqrt{x}+1\), \(u(x)=4x^5\)

- Work from the innermost operation to the outermost operation. - In part b, the inner function must contain the entire radicand. - Check by composing your functions.

1. One three-function decomposition is \(k(x)=x^2\), \(g(x)=\cos(x)+2\), and \(f(x)=\sqrt{x}\). Composing gives \(f(g(k(x)))=\sqrt{\cos(x^2)+2}\). 2. For a two-function decomposition with a square-root outer function, choose \(v(x)=\cos(x^2)+2\) and \(u(x)=\sqrt{x}\).

a) One possible answer is \(k(x)=x^2\), \(g(x)=\cos(x)+2\), and \(f(x)=\sqrt{x}\). b) \(v(x)=\cos(x^2)+2\) and \(u(x)=\sqrt{x}\)

- Identify the operation performed first; that is the inner function. - Identify the final operation; that is the outer function. - Check your answer by composing the two functions.

1. For part a, choose \(v(x)=3x-4\) and \(u(x)=\ln(x)\). 2. For part b, choose \(v(x)=\cos(x)\) and \(u(x)=\sqrt{x}\). 3. For part c, choose \(v(x)=e^x+2\) and \(u(x)=x^5\). 4. For part d, choose \(v(x)=\frac{1}{x}\) and \(u(x)=e^x\).

a) \(v(x)=3x-4\), \(u(x)=\ln(x)\) b) \(v(x)=\cos(x)\), \(u(x)=\sqrt{x}\) c) \(v(x)=e^x+2\), \(u(x)=x^5\) d) \(v(x)=\frac{1}{x}\), \(u(x)=e^x\)

- Work from the innermost operation to the outermost operation. - The function applied directly to \(x\) is \(w\). - Recall that \(\sin^4(x)=(\sin(x))^4\).

1. For part a, choose \(w(x)=5x+2\), \(v(x)=x^3\), and \(u(x)=\frac{1}{x}\). 2. For part b, choose \(w(x)=x^2\), \(v(x)=\cos(x)\), and \(u(x)=\sqrt{x}\). 3. For part c, choose \(w(x)=x-\pi\), \(v(x)=\sin(x)\), and \(u(x)=x^4\).

a) \(w(x)=5x+2\), \(v(x)=x^3\), \(u(x)=\frac{1}{x}\) b) \(w(x)=x^2\), \(v(x)=\cos(x)\), \(u(x)=\sqrt{x}\) c) \(w(x)=x-\pi\), \(v(x)=\sin(x)\), \(u(x)=x^4\)

- Identify the outermost structure first. - Then separate the remaining inner operations. - Check your decomposition by recomposing the functions.

1. For part a, choose \(w(x)=2x\), \(v(x)=\sin(x)\), and \(u(x)=\sqrt[3]{x}\). 2. For part b, choose \(w(x)=x^2+4x\), \(v(x)=\sqrt{x}\), and \(u(x)=\frac{1}{x}\). 3. For part c, choose \(w(x)=\cos(x)\), \(v(x)=x+1\), and \(u(x)=x^5\).

a) \(w(x)=2x\), \(v(x)=\sin(x)\), \(u(x)=\sqrt[3]{x}\) b) \(w(x)=x^2+4x\), \(v(x)=\sqrt{x}\), \(u(x)=\frac{1}{x}\) c) \(w(x)=\cos(x)\), \(v(x)=x+1\), \(u(x)=x^5\)

- Identify the expression evaluated first. - The final operation usually defines the outer function. - Check by composing the two functions.

1. The expression inside the parentheses is evaluated first, so choose \(v(x)=x+\frac{1}{x}\). 2. The final operation is cubing, so choose \(u(x)=x^3\). 3. Then \(u(v(x))=\left(x+\frac{1}{x}\right)^3=f(x)\).

One possible answer is \(v(x)=x+\frac{1}{x}\) and \(u(x)=x^3\).

- More than one decomposition may be correct. - Try choosing the sine function as the inner function. - Check by recomposing the functions.

1. One natural choice is \(v(x)=\sin(x)\). 2. The outer function must square its input, add \(1\), and take the square root, so choose \(u(x)=\sqrt{1+x^2}\). 3. Then \(u(v(x))=\sqrt{1+(\sin(x))^2}=f(x)\). Other valid decompositions are possible.

One possible answer is \(v(x)=\sin(x)\) and \(u(x)=\sqrt{1+x^2}\).

- Evaluate the inner function first. - Use the inner output as the x-coordinate for the outer graph. - Pay attention to the order of composition.

1. For \(h(x)=f(g(x))\), read \(g(x)\) first and then use that result as the input of \(f\). The graph gives \(g(0)=2\), \(g(2)=0\), and \(g(4)=-2\). Therefore, \(h(0)=f(2)=0\), \(h(2)=f(0)=-2\), and \(h(4)=f(-2)=0\). 2. For \(k(x)=g(f(x))\), read \(f(x)\) first. The graph gives \(f(-2)=0\), \(f(0)=-2\), and \(f(2)=0\). Therefore, \(k(-2)=g(0)=2\), \(k(0)=g(-2)=4\), and \(k(2)=g(0)=2\).

a) \(0\) b) \(-2\) c) \(0\) d) \(2\) e) \(4\) f) \(2\)

- Evaluate the inner function first. - Use the first output as the x-value on the second graph. - Composition is generally order-dependent.

1. From the graph, \(g(2)=0\) and \(f(0)=3\). Therefore, \(h_1(2)=f(g(2))=3\). 2. From the graph, \(f(2)=1\) and \(g(1)=-0.5\). Therefore, \(h_2(2)=g(f(2))=-0.5\). 3. Since \(3\ne-0.5\), the two compositions are not equal at \(x=2\).

1) \(h_1(2)=3\) 2) \(h_2(2)=-0.5\) \(h_1(2)\ne h_2(2)\)

- Locate the peak of the displayed graph. - Recall how \(\ln(u)\) changes as a positive input \(u\) increases. - A strictly increasing outer function preserves comparisons between nearby function values.

1. The graph shows a local maximum of \(h\) at \(x_0=4\). 2. The natural logarithm is strictly increasing on its domain. Therefore, it preserves the order of positive function values: if \(h(x_1)<h(x_2)\), then \(\ln(h(x_1))<\ln(h(x_2))\). 3. Near \(x=4\), the values of \(h\) are no greater than \(h(4)\). Applying the increasing logarithm preserves that comparison, so nearby values of \(f\) are no greater than \(f(4)\). Thus \(f\) also has a local maximum at \(x_0=4\).

\(x_0=4\). Because \(\ln(u)\) is strictly increasing for \(u>0\), it preserves the location and type of the local maximum.

- Track the unit entering and leaving each function. - Ask which rule can accept the original ride-time input. - The inner function must produce the kind of quantity that the outer function accepts.

1. The starting quantity is in minutes, so it can be used as an input to \(c\). 2. The output \(c(t)\) is in dollars, which is exactly the input unit required by \(d\). 3. Therefore, \(d(c(t))\) is meaningful. The expression \(c(d(t))\) is not meaningful from the stated units because \(d\) cannot take the original minute input.

\(d(c(t))\). Minutes enter \(c\), dollars come out of \(c\), and those dollars can then enter \(d\).

- Start with the innermost function and use the first table. - The output from the first lookup becomes the input for the second lookup. - Check the units as the quantity moves from one table to the next.

1. From the first table, \(p(4)=35\) test runs. 2. Use that output as the input of \(s\). From the second table, \(s(35)=20\) points. 3. Therefore, \(s(p(4))=20\), meaning that \(4\) hours of practice corresponds to a score of \(20\) points in this two-stage model.

\(s(p(4))=20\) points. In the model, \(4\) hours of practice leads to a score of \(20\) points.

- Identify the domain of each original function first. - For a sum, both functions must be defined. - For a composition, determine which function is applied first. - A rational expression is undefined when its denominator is zero.

1. The domains of the original functions are \(D_f=\mathbb{R}\) and \(D_g=\mathbb{R}\setminus\{-3\}\). 2. For the sum, \((f+g)(x)=x^2-1+\frac{2}{x+3}\). Both functions must be defined, so the domain is \(\mathbb{R}\setminus\{-3\}\). 3. For \(g\circ f\), \((g\circ f)(x)=\frac{2}{(x^2-1)+3}=\frac{2}{x^2+2}\). The denominator is positive for every real \(x\), so the domain is \(\mathbb{R}\). 4. For \(f\circ g\), \((f\circ g)(x)=\left(\frac{2}{x+3}\right)^2-1=\frac{4}{(x+3)^2}-1\). The only restriction is \(x\ne-3\).

a) \((f+g)(x)=x^2-1+\frac{2}{x+3}\), \(D=\mathbb{R}\setminus\{-3\}\) b) \((g\circ f)(x)=\frac{2}{x^2+2}\), \(D=\mathbb{R}\) c) \((f\circ g)(x)=\frac{4}{(x+3)^2}-1\), \(D=\mathbb{R}\setminus\{-3\}\)

- The input of a natural logarithm must be positive. - Pay attention to the order of the functions in each composition. - Solve the inequality that makes the logarithm input positive. - Squaring the output of a logarithm adds no new domain restriction.

1. The original domains are \(D_u=(0, \infty)\) and \(D_v=\mathbb{R}\). 2. For the product, \((uv)(x)=(x^2-4)\ln(x)\). Both factors must be defined, so the domain is \((0, \infty)\). 3. For \(u\circ v\), \((u\circ v)(x)=\ln(x^2-4)\). The logarithm requires \(x^2-4>0\), so the domain is \((-\infty, -2)\cup(2, \infty)\). 4. For \(v\circ u\), \((v\circ u)(x)=(\ln(x))^2-4\). The logarithm requires \(x>0\), and there are no further restrictions, so the domain is \((0, \infty)\).

a) \((uv)(x)=(x^2-4)\ln(x)\), \(D=(0, \infty)\) b) \((u\circ v)(x)=\ln(x^2-4)\), \(D=(-\infty, -2)\cup(2, \infty)\) c) \((v\circ u)(x)=(\ln(x))^2-4\), \(D=(0, \infty)\)

- Determine the domain of the inner function first. - The output of the inner function must be in the domain of the outer function. - A denominator cannot be zero. - A square-root radicand must be nonnegative.

1. For part a, \(f(x)=u(v(x))=\frac{2}{x^2-4}\). The denominator cannot equal zero, so \(x\ne-2\) and \(x\ne2\). Thus \(D_f=\mathbb{R}\setminus\{-2, 2\}\). 2. Also, \(g(x)=v(u(x))=\left(\frac{2}{x-4}\right)^2=\frac{4}{(x-4)^2}\). The inner function is undefined at \(x=4\), so \(D_g=\mathbb{R}\setminus\{4\}\). 3. For part b, \(f(x)=u(v(x))=\sqrt{x}-5\). The square root requires \(x\ge0\), so \(D_f=[0, \infty)\). 4. Also, \(g(x)=v(u(x))=\sqrt{x-5}\). The radicand must be nonnegative, so \(x\ge5\) and \(D_g=[5, \infty)\).

a) \(f(x)=\frac{2}{x^2-4}\), \(D_f=\mathbb{R}\setminus\{-2, 2\}\); \(g(x)=\frac{4}{(x-4)^2}\), \(D_g=\mathbb{R}\setminus\{4\}\) b) \(f(x)=\sqrt{x}-5\), \(D_f=[0, \infty)\); \(g(x)=\sqrt{x-5}\), \(D_g=[5, \infty)\)

- Substitute the entire inner function into the outer function. - Two linear functions are identical when their corresponding coefficients are equal. - Use a simple value of \(c\) for the counterexample.

1. The compositions are \(h(x)=f(g(x))=2(3x+c)+5=6x+2c+5\) and \(k(x)=g(f(x))=3(2x+5)+c=6x+15+c\). 2. The x-coefficients already agree. Equating the constant terms gives \(2c+5=15+c\), so \(c=10\). 3. For example, if \(c=0\), then \(h(x)=6x+5\) and \(k(x)=6x+15\). These functions are not equal, so the compositions are not commutative for this value. The same conclusion holds for every \(c\ne10\).

a) \(h(x)=6x+2c+5\) and \(k(x)=6x+15+c\) b) \(c=10\) c) For example, when \(c=0\), \(h(x)=6x+5\ne6x+15=k(x)\).

- Check the domain of the inner function first. - The input of a logarithm must be positive. - A rational expression is undefined when its denominator is zero.

1. For \(f=u\circ v\), \(f(x)=\ln\left(\frac{1}{x^2-4}\right)\). The logarithm input must be positive, so \(\frac{1}{x^2-4}>0\). This requires \(x^2-4>0\), or \(|x|>2\). Thus \(D_f=(-\infty, -2)\cup(2, \infty)\). 2. For \(g=v\circ u\), \(g(x)=\frac{1}{(\ln(x))^2-4}\). First, \(x>0\). Also, the denominator cannot be zero, so \((\ln(x))^2\ne4\). Therefore, \(x\ne e^{-2}\) and \(x\ne e^2\). 3. Hence \(D_g=(0, e^{-2})\cup(e^{-2}, e^2)\cup(e^2, \infty)\).

\(f(x)=\ln\left(\frac{1}{x^2-4}\right)\), \(D_f=(-\infty, -2)\cup(2, \infty)\) \(g(x)=\frac{1}{(\ln(x))^2-4}\), \(D_g=(0, e^{-2})\cup(e^{-2}, e^2)\cup(e^2, \infty)\)

- A square-root radicand must be nonnegative. - A denominator cannot be zero. - Use a sign chart for the rational inequality. - Begin with the domain of the inner function.

1. For \(f=u\circ v\), \(f(x)=\sqrt{\frac{x+3}{x-1}}\). The rational expression must be defined and nonnegative. A sign analysis at the critical values \(-3\) and \(1\) gives \(D_f=(-\infty, -3]\cup(1, \infty)\). 2. For \(g=v\circ u\), \(g(x)=\frac{\sqrt{x}+3}{\sqrt{x}-1}\). The square root requires \(x\ge0\), and the denominator requires \(\sqrt{x}\ne1\), so \(x\ne1\). 3. Therefore, \(D_g=[0, 1)\cup(1, \infty)\).

\(f(x)=\sqrt{\frac{x+3}{x-1}}\), \(D_f=(-\infty, -3]\cup(1, \infty)\) \(g(x)=\frac{\sqrt{x}+3}{\sqrt{x}-1}\), \(D_g=[0, 1)\cup(1, \infty)\)

- Follow the operations from the inside outward. - Find the smallest possible value of the logarithm input. - Determine how the reciprocal changes as its positive denominator increases.

1. One decomposition follows the order of operations: \(h(x)=x^2+e\), \(g(x)=\ln(x)\), and \(f(x)=\frac{4}{x}\). Then \((f\circ g\circ h)(x)=\frac{4}{\ln(x^2+e)}\). 2. Since \(x^2\ge0\), \(x^2+e\ge e\), so \(\ln(x^2+e)\ge1\). 3. The denominator takes every value in \([1, \infty)\). Therefore, \(k(x)\) is positive, has maximum value \(4\), and approaches \(0\) without reaching it. The range is \((0, 4]\).

a) One possible decomposition is \(h(x)=x^2+e\), \(g(x)=\ln(x)\), and \(f(x)=\frac{4}{x}\). b) \((0, 4]\)

- In \((a\circ b)(x)\), evaluate \(b\) first. - Use the first output as the input for the second table lookup.

1. For \(f(x)=a(b(x))\), read the value of \(b(x)\) first and then use it as the input of \(a\). This gives \(f(-3)=1\), \(f(-2)=-2\), \(f(-1)=-3\), \(f(0)=0\), and \(f(1)=-1\). 2. For \(g(x)=b(a(x))\), read the value of \(a(x)\) first and then use it as the input of \(b\). This gives \(g(-3)=-3\), \(g(-2)=0\), \(g(-1)=-2\), \(g(0)=-1\), and \(g(1)=1\).

<table> <tr> <td>\(x\)</td> <td>\(-3\)</td> <td>\(-2\)</td> <td>\(-1\)</td> <td>\(0\)</td> <td>\(1\)</td> </tr> <tr> <td>\(f(x)\)</td> <td>\(1\)</td> <td>\(-2\)</td> <td>\(-3\)</td> <td>\(0\)</td> <td>\(-1\)</td> </tr> <tr> <td>\(g(x)\)</td> <td>\(-3\)</td> <td>\(0\)</td> <td>\(-2\)</td> <td>\(-1\)</td> <td>\(1\)</td> </tr> </table>

- Evaluate the inner function first. - In part b, compute both compositions for each listed input and compare them.

1. Since \(v(4)=3\) and \(u(3)=4\), \(u(v(4))=4\). 2. Compare the two compositions at each input. They are equal at \(x=1\), where both values are \(1\), and at \(x=3\), where both values are \(3\). 3. At \(x=2\), \(x=4\), and \(x=5\), the two composition values are different. Therefore, the solution set is \(\{1, 3\}\).

a) \(4\) b) \(x\in\{1, 3\}\)

- Treat the entire radicand as one expression in part a. - Break the inner expression into smaller steps for part b. - In part c, replace the repeated inner expression with a temporary variable.

1. For part a, choose \(v(x)=e^{2x}+1\) and \(u(x)=\sqrt{x}\). 2. One three-function decomposition is \(r(x)=2x\), \(q(x)=e^x+1\), and \(p(x)=\sqrt{x}\). 3. Since \(v(x)=e^{2x}+1\) and \(g(x)=\frac{1}{v(x)}\), choose \(w(x)=\frac{1}{x}\).

a) \(v(x)=e^{2x}+1\), \(u(x)=\sqrt{x}\) b) One possible answer is \(r(x)=2x\), \(q(x)=e^x+1\), and \(p(x)=\sqrt{x}\). c) \(w(x)=\frac{1}{x}\)

- Track what happens to the highest power under substitution. - Use \((x^m)^n=x^{mn}\). - Expand the composition in part 3.

1. Write the leading terms as \(p(x)=a_nx^n+\cdots\) and \(q(x)=b_mx^m+\cdots\), where \(a_n\ne0\) and \(b_m\ne0\). The leading term of \(p(q(x))\) is \(a_n(b_mx^m)^n=a_nb_m^nx^{mn}\). Therefore, the degree is \(mn\). 2. Eleni's claim is false. If both functions have degree \(2\), their composition has degree \(2\cdot2=4\), not degree \(2\). 3. \(h(x)=(x^2-2)^2-2=x^4-4x^2+2\), so its degree is \(4\).

1) \(mn\) 2) The claim is false; the composition has degree \(4\). 3) \(h(x)=x^4-4x^2+2\), degree \(4\)

- Treat the expression as an outer function applied to an inner function rather than analyzing both layers at once. - Determine every possible output of \(v(x)=2+\sin(x)\) before checking whether those outputs are legal inputs to \(u\). - For the final range, ask what an increasing outer function does to the endpoints of the inner range.

1. Since \(-1\le\sin(x)\le1\), adding \(2\) gives \(1\le v(x)\le3\). Thus, the range of \(v\) is \([1,3]\). 2. The outer function \(u(t)=\ln(t)\) requires \(t>0\). Every output of \(v\) lies in \([1,3]\), so every real input \(x\) produces an allowed input for \(u\). Therefore, the domain of \(f=u\circ v\) is \(\mathbb{R}\). 3. Because \(u\) is increasing, it maps the interval \([1,3]\) to \([\ln(1),\ln(3)]\). Hence, the range of \(f\) is \([0,\ln(3)]\).

a) \([1,3]\) b) \(\mathbb{R}\) c) \([0,\ln(3)]\)

- Substitute the entire inner function into the outer function. - A denominator cannot be zero. - Check the inner function domain before considering the outer function.

1. \(f(x)=u(v(x))=\frac{3}{(x^2+1)-4}=\frac{3}{x^2-3}\). The denominator is zero at \(x=\pm\sqrt{3}\), so \(D_f=\mathbb{R}\setminus\{-\sqrt{3}, \sqrt{3}\}\). 2. \(g(x)=v(u(x))=\left(\frac{3}{x-4}\right)^2+1=\frac{9}{(x-4)^2}+1\). The inner function is undefined at \(x=4\), so \(D_g=\mathbb{R}\setminus\{4\}\).

\(f(x)=\frac{3}{x^2-3}\), \(D_f=\mathbb{R}\setminus\{-\sqrt{3}, \sqrt{3}\}\) \(g(x)=\frac{9}{(x-4)^2}+1\), \(D_g=\mathbb{R}\setminus\{4\}\)

- The input of a logarithm must be positive. - Check the inner function domain first. - A polynomial outer function accepts every real input.

1. \(f(x)=u(v(x))=\ln(4-x^2)\). The logarithm requires \(4-x^2>0\), so \(-2<x<2\). Therefore, \(D_f=(-2, 2)\). 2. \(g(x)=v(u(x))=4-(\ln(x))^2\). The logarithm requires \(x>0\), while the polynomial outer function adds no restrictions. Therefore, \(D_g=(0, \infty)\).

\(f(x)=\ln(4-x^2)\), \(D_f=(-2, 2)\) \(g(x)=4-(\ln(x))^2\), \(D_g=(0, \infty)\)

- Substitute in the correct order. - A square-root radicand must be nonnegative. - A denominator cannot be zero. - Use a sign chart for the rational radicand.

1. \(f(x)=u(v(x))=\sqrt{\frac{2}{x-1}+2}=\sqrt{\frac{2x}{x-1}}\). The radicand must be nonnegative and \(x\ne1\). A sign analysis gives \(D_f=(-\infty, 0]\cup(1, \infty)\). 2. \(g(x)=v(u(x))=\frac{2}{\sqrt{x+2}-1}\). The square root requires \(x\ge-2\), and the denominator requires \(x\ne-1\). Thus \(D_g=[-2, -1)\cup(-1, \infty)\).

\(f(x)=\sqrt{\frac{2x}{x-1}}\), \(D_f=(-\infty, 0]\cup(1, \infty)\) \(g(x)=\frac{2}{\sqrt{x+2}-1}\), \(D_g=[-2, -1)\cup(-1, \infty)\)

- Test whether \(g(-x)=g(x)\). - First determine where the exponent \(-x^2+1\) is greatest. - Analyze the exponent as \(|x|\) becomes large.

1. \(g(-x)=2^{-(-x)^2+1}=2^{-x^2+1}=g(x)\), so \(g\) is even and its graph is symmetric about the y-axis. 2. Because the base \(2\) is greater than \(1\), \(g(x)\) is largest when the exponent \(-x^2+1\) is largest. This occurs at \(x=0\), where \(g(0)=2\). The highest point is \((0, 2)\). 3. As \(x\to\infty\) or \(x\to-\infty\), the exponent \(-x^2+1\to-\infty\). Therefore, \(g(x)\to0\) in both directions. 4. Geometrically, the x-axis, \(y=0\), is a horizontal asymptote on both ends of the graph.

a) Yes; \(g(-x)=g(x)\), so the graph is symmetric about the y-axis. b) \((0, 2)\) c) \(g(x)\to0\) as \(x\to\infty\) and as \(x\to-\infty\); \(y=0\) is a horizontal asymptote.

- Work from the inside outward. - Check the reciprocal and square-root restrictions when finding each domain. - Substitute \(x=1\) only after simplifying the compositions.

1. \(g(x)=u(v(w(x)))=\sqrt{\frac{1}{x^2}+2}\). The reciprocal requires \(x\ne0\), and the radicand is positive for every such \(x\). Therefore, \(D_g=\mathbb{R}\setminus\{0\}\). 2. \(h(x)=w(v(u(x)))=\frac{1}{(\sqrt{x})^2+2}=\frac{1}{x+2}\). The original composition requires \(x\ge0\), so \(D_h=[0, \infty)\). 3. \(g(1)=\sqrt{3}\) and \(h(1)=\frac{1}{3}\). The values differ because changing the order of function composition generally changes the result.

a) \(g(x)=\sqrt{\frac{1}{x^2}+2}\), \(D_g=\mathbb{R}\setminus\{0\}\); \(h(x)=\frac{1}{x+2}\), \(D_h=[0, \infty)\) b) \(g(1)=\sqrt{3}\) and \(h(1)=\frac{1}{3}\)

- The output of \(v\) must be nonnegative. - Write and solve an inequality for each part. - Use a sign chart for the rational inequality. - Preserve any exclusions from the original domain.

1. Each composition requires \(v(x)\ge0\). 2. For part a, \(5-2x\ge0\) gives \(x\le\frac{5}{2}\). The domain is \((-\infty, \frac{5}{2}]\). 3. For part b, \(x^2-16\ge0\) gives \(x\le-4\) or \(x\ge4\). The domain is \((-\infty, -4]\cup[4, \infty)\). 4. For part c, solve \(\frac{x+4}{x-2}\ge0\) using the critical values \(-4\) and \(2\). The solution is \(x\le-4\) or \(x>2\). Thus the domain is \((-\infty, -4]\cup(2, \infty)\).

a) \((-\infty, \frac{5}{2}]\) b) \((-\infty, -4]\cup[4, \infty)\) c) \((-\infty, -4]\cup(2, \infty)\)

- The input of a logarithm must be positive. - Use logarithms to solve the exponential inequality. - A fraction with positive numerator is positive when its denominator is positive.

1. Each composition requires \(v(x)>0\). 2. For part a, \(e^x-5>0\) gives \(e^x>5\), so \(x>\ln(5)\). The domain is \((\ln(5), \infty)\). 3. For part b, \(\frac{1}{x^2-1}>0\). Since the numerator is positive, the denominator must be positive: \(x^2-1>0\). Thus \(x<-1\) or \(x>1\), and the domain is \((-\infty, -1)\cup(1, \infty)\).

a) \((\ln(5), \infty)\) b) \((-\infty, -1)\cup(1, \infty)\)

- Evaluate the inner graph before the outer graph. - For a composition to equal zero, the inner output must be a zero of the outer function. - Use the vertex and intercepts to identify the quadratic equation. - Use two points to identify the linear equation.

1. Since \(v(0)=0\) and \(u(0)=-1\), \(f(0)=u(v(0))=-1\). Since \(u(2)=0\) and \(v(0)=0\), \(g(2)=v(u(2))=0\). 2. For \(f(x)=0\), the value \(v(x)\) must equal a zero of \(u\). The graph of \(u\) has zeros \(-2\) and \(2\). Since \(v(-4)=-2\) and \(v(4)=2\), the zeros of \(f\) are \(-4\) and \(4\). 3. For \(g(x)=0\), \(u(x)\) must equal the zero of \(v\), which is \(0\). Thus the zeros of \(g\) are the zeros of \(u\): \(-2\) and \(2\). 4. The graphs give \(u(x)=0.25x^2-1\) and \(v(x)=0.5x\). Therefore, \(f(x)=0.25(0.5x)^2-1=0.0625x^2-1\), and \(g(x)=0.5(0.25x^2-1)=0.125x^2-0.5\).

a) \(f(0)=-1\) and \(g(2)=0\) b) Zeros of \(f\): \(x=-4, 4\); zeros of \(g\): \(x=-2, 2\) c) \(u(x)=0.25x^2-1\), \(v(x)=0.5x\), \(f(x)=0.0625x^2-1\), and \(g(x)=0.125x^2-0.5\)

- Write an equation for each possible composition first. - Distinguish horizontal shifts from vertical shifts. - Compare the vertices of the two possible parabolas.

1. The four possible equations are \(u(x)=\cos(x-1)\), \(v(x)=\cos(x)-1\), \(w(x)=(x-1)^2\), and \(z(x)=x^2-1\). 2. Graph p is a cosine curve shifted \(1\) unit right, with a maximum at \(x=1\), so it represents \(u(x)=\cos(x-1)\). 3. Graph q is a cosine curve shifted \(1\) unit down, with range \([-2, 0]\), so it represents \(v(x)=\cos(x)-1\). 4. Graph r is an upward-opening parabola with vertex \((0, -1)\), so it represents \(z(x)=x^2-1\), not \(w(x)=(x-1)^2\).

Graph p: \(u(x)=g(h(x))=\cos(x-1)\); Graph q: \(v(x)=h(g(x))=\cos(x)-1\); Graph r: \(z(x)=h(f(x))=x^2-1\); unused: \(w(x)=f(h(x))=(x-1)^2\)

- Use symmetry and the vertex to identify the parabola. - Compare y-intercepts and horizontal asymptotes for the exponential graph. - Write equations for likely compositions and compare their features.

1. Graph p is a downward-opening parabola with vertex \((0, 2)\) and y-axis symmetry. The composition \(w(v(x))=-x^2+2\) has exactly these features, so graph p represents \(w\circ v\). 2. Graph q approaches the horizontal asymptote \(y=2\) as \(x\to-\infty\), passes through \((0, 1)\), and decreases without bound as \(x\to\infty\). The composition \(w(u(x))=-e^x+2\) has these features, so graph q represents \(w\circ u\).

\(p: (w\circ v)(x)=-x^2+2\); \(q: (w\circ u)(x)=-e^x+2\)

- Read the inner-function values from the graph first. - Consider what happens to \(e^u\) as \(u\to-\infty\). - A strictly increasing outer function preserves the location of maxima.

1. The graph gives \(f(0)=1\) and \(f(2)=0\). Therefore, \(g(0)=e^1=e\) and \(g(2)=e^0=1\). 2. The graph of \(f\) is a downward-opening parabola, so \(f(x)\to-\infty\) as \(x\to\pm\infty\). Therefore, \(g(x)=e^{f(x)}\to0\) in both directions. 3. Because the exponential function is strictly increasing, \(g\) reaches its maximum at the same x-value as \(f\). The graph of \(f\) has maximum value \(1\) at \(x=0\), so the maximum point of \(g\) is \((0, e)\).

a) \(g(0)=e\) and \(g(2)=1\) b) \(\lim_{x\to\infty}g(x)=0\) and \(\lim_{x\to-\infty}g(x)=0\) c) Maximum point: \((0, e)\)

- The input of a logarithm must be positive. - Identify where the graph of \(f\) lies above the x-axis. - A strictly increasing outer function preserves monotonicity.

1. The logarithm requires \(f(x)>0\). From the graph, \(f\) is positive only for \(-1<x<3\). The point \(x=-1\) is a vertical asymptote and \(x=3\) is a zero, so neither endpoint is included. Thus \(D_h=(-1, 3)\). 2. The natural logarithm is strictly increasing, so it preserves the monotonic behavior of positive inputs. The graph of \(f\) is strictly decreasing on \((-1, 3)\). Therefore, \(h\) is strictly decreasing throughout its domain.

Domain: \((-1, 3)\) Monotonic behavior: strictly decreasing on \((-1, 3)\)

- Evaluate the inner function first. - Work backward from the output of the outer function when solving an equation. - Changes inside the exponent produce horizontal shifts; changes outside produce vertical shifts.

1. Since \(v(3)=1\) and \(u(1)=2\), \(f(3)=2\). Since \(u(1)=2\) and \(v(2)=0\), \(g(1)=0\). 2. The equation \(f(x)=1\) means \(u(v(x))=1\). Since \(u(0)=1\), solve \(v(x)=0\), which gives \(x=2\). The equation \(g(x)=0\) means \(v(u(x))=0\). Since \(v(2)=0\), solve \(u(x)=2\), which gives \(x=1\). 3. Since \(f(x)=2^{x-2}\), its graph is the graph of \(u\) shifted \(2\) units right. Since \(g(x)=2^x-2\), its graph is the graph of \(u\) shifted \(2\) units down.

a) \(f(3)=2\) and \(g(1)=0\) b) \(f(x)=1\) at \(x=2\); \(g(x)=0\) at \(x=1\) c) \(f\): shift \(2\) units right; \(g\): shift \(2\) units down

- Write an equation for each composition. - Identify reflections and horizontal or vertical shifts. - Compare y-intercepts and end behavior.

1. \(f(g(x))=2^{-x}\) is a decreasing exponential function with y-intercept \(1\), so it is graph p. 2. \(g(f(x))=-2^x\) is the graph of \(2^x\) reflected across the x-axis, so it is graph q. 3. \(f(h(x))=2^{x-3}\) is the graph of \(2^x\) shifted \(3\) units right, so it is graph r. 4. \(h(f(x))=2^x-3\) is the graph of \(2^x\) shifted \(3\) units down, so it is graph s.

p: \(k_a(x)=f(g(x))\); q: \(k_b(x)=g(f(x))\); r: \(k_c(x)=f(h(x))\); s: \(k_d(x)=h(f(x))\)

- Check whether each graph can have negative outputs. - Look for y-axis symmetry and sharp corners. - Identify vertical shifts from the basic graphs.

1. \(f(g(x))=|\sin(x)|\) is periodic and never negative, so it is graph p. 2. \(g(f(x))=\sin(|x|)\) is symmetric about the y-axis and has a corner at the origin, so it is graph q. 3. \(h(f(x))=|x|-2\) is a V-shaped graph with vertex \((0, -2)\), so it is graph r. 4. \(h(g(x))=\sin(x)-2\) is a sine curve shifted \(2\) units down, so it is graph s.

p) \(u_1(x)=f(g(x))\); q) \(u_2(x)=g(f(x))\); r) \(u_3(x)=h(f(x))\); s) \(u_4(x)=h(g(x))\)

- Identify where the graph of \(f\) lies above the x-axis. - Recall the behavior of \(\ln(u)\) as \(u\to0^+\) and as \(u\to\infty\). - Consider the composition of two increasing functions.

1. From the graph, \(f(x)=0.5x+1\). The logarithm requires \(f(x)>0\), so \(0.5x+1>0\), which gives \(x>-2\). Thus, the domain is \((-2, \infty)\). 2. As \(x\to-2^+\), \(f(x)\to0^+\), so \(g(x)\to-\infty\). As \(x\to\infty\), \(f(x)\to\infty\), so \(g(x)\to\infty\). 3. The linear function \(f\) is strictly increasing, and \(\ln(x)\) is strictly increasing. Therefore, their composition \(g\) is strictly increasing on \((-2, \infty)\).

a) \((-2, \infty)\) b) \(\lim_{x\to-2^+}g(x)=-\infty\) and \(\lim_{x\to\infty}g(x)=\infty\) c) \(g\) is strictly increasing.

- Find where the graph of \(f\) is above the x-axis. - Read the monotonicity of \(f\) on those intervals. - An increasing outer function preserves the monotonicity of the inner function.

1. The logarithm requires \(f(x)>0\). From the graph, \(f(x)>0\) for \(x<-1\) and \(x>1\). Thus, the domain is \((-\infty, -1)\cup(1, \infty)\). 2. The graph shows that \(f\) is strictly decreasing on both \((-\infty, -1)\) and \((1, \infty)\). 3. Since the natural logarithm is strictly increasing, composing with it preserves the direction of monotonicity. Therefore, \(g\) is strictly decreasing on each domain interval.

Domain: \((-\infty, -1)\cup(1, \infty)\) The function is strictly decreasing on \((-\infty, -1)\) and on \((1, \infty)\).

- Replace the input of \(f\) by the unknown expression \(g(x)\). - Work backward from the required output while respecting the square root's nonnegative output. - In the verification, pay attention to the difference between \(\sqrt{z^2}\) and \(z\) for arbitrary real \(z\).

1. The equation \(f(g(x))=x+4\) means \(\sqrt{g(x)+2}=x+4\). 2. For \(x\ge-4\), the right side is nonnegative, so squaring gives \(g(x)+2=(x+4)^2\). 3. Therefore, \(g(x)=(x+4)^2-2\). 4. Verify: \(f(g(x))=\sqrt{(x+4)^2}=|x+4|\). Because \(x\ge-4\), \(|x+4|=x+4\), as required.

\(g(x)=(x+4)^2-2\) for \(x\ge-4\). The restriction ensures \(\sqrt{(x+4)^2}=|x+4|=x+4\).

- Find where the denominator equals zero. - Use the range of the cosine function to find the range of the denominator. - Analyze the reciprocal separately for negative and positive denominator values.

1. The denominator is zero when \(2\cos(x)-1=0\), or \(\cos(x)=\frac{1}{2}\). Thus \(x=2\pi k\pm\frac{\pi}{3}\), where \(k\in\mathbb{Z}\), must be excluded from the domain. 2. Since \(-1\le\cos(x)\le1\), the denominator \(u=2\cos(x)-1\) takes every value in \([-3, 1]\), except \(0\) when used in \(g\). 3. For \(u\in[-3, 0)\), \(\frac{2}{u}\in(-\infty, -\frac{2}{3}]\). For \(u\in(0, 1]\), \(\frac{2}{u}\in[2, \infty)\). 4. Therefore, the range is \((-\infty, -\frac{2}{3}]\cup[2, \infty)\).

Domain: \(\mathbb{R}\setminus\left\{2\pi k\pm\frac{\pi}{3}: k\in\mathbb{Z}\right\}\) Range: \((-\infty, -\frac{2}{3}]\cup[2, \infty)\)

- Apply the domain restriction of the logarithm. - Use exponential and logarithm properties to simplify. - Simplifying a formula does not restore inputs excluded by the original composition.

1. \(f(x)=u(v(x))=\ln((e^{x+1})^2-1)=\ln(e^{2x+2}-1)\). The logarithm requires \(e^{2x+2}-1>0\), which is equivalent to \(x>-1\). Thus \(D_f=(-1, \infty)\). 2. \(g(x)=v(u(x))=e^{\ln(x^2-1)+1}=e(x^2-1)\). The original composition still requires \(x^2-1>0\), even though the formula simplifies. Therefore, \(D_g=(-\infty, -1)\cup(1, \infty)\).

\(f(x)=\ln(e^{2x+2}-1)\), \(D_f=(-1, \infty)\) \(g(x)=e(x^2-1)\), \(D_g=(-\infty, -1)\cup(1, \infty)\)

- The vertical shift determines the horizontal asymptote. - Choose an even polynomial exponent for y-axis symmetry. - Make the exponent approach \(-\infty\) at both ends. - Use the given point to determine the remaining coefficient.

1. Choose \(c=1\) so that \(y=1\) can be the horizontal asymptote. 2. Choose the even polynomial \(g(x)=-x^2\). Then \(g(x)\to-\infty\) as \(x\to\pm\infty\), so \(e^{g(x)}\to0\), and the resulting function is even. 3. Since \(g(0)=0\), the condition \(h(0)=-1\) gives \(a+1=-1\), so \(a=-2\). 4. One valid function is \(h(x)=-2e^{-x^2}+1\). It is even, approaches \(1\) as \(x\to\pm\infty\), and satisfies \(h(0)=-1\). 5. Since \(e^{-x^2}\le1\) for every real \(x\), \(-2e^{-x^2}+1\ge-1\), with equality at \(x=0\). Therefore, \((0,-1)\) is a local minimum (in fact, a global minimum).

One valid choice is \(a=-2\), \(c=1\), and \(g(x)=-x^2\), giving \(h(x)=-2e^{-x^2}+1\).

- Use the marked point on the red graph to determine the coefficient in \(h(x)=ax^2\). - Compare \(h(x)\) with a constant multiple of \(f(x)\). - Rewrite the coefficient as a square to express \(h\) in the form \(f(cx)\).

1. The red graph passes through \((1,4)\). If \(h(x)=ax^2\), then \(4=a(1)^2\), so \(h(x)=4x^2\). 2. Since \(h(x)=4f(x)\), the graph is a vertical stretch of \(f\) by a factor of \(4\). 3. Also, \(h(x)=f(2x)=(2x)^2=4x^2\). Replacing \(x\) with \(2x\) produces a horizontal compression by a factor of \(\frac{1}{2}\).

\(h(x)=4x^2\) Vertical stretch factor: \(4\) Horizontal compression factor: \(\frac{1}{2}\)

- First recover the line rule on each visible segment before applying either transformation. - For an input transformation, transform the interval conditions as well as the formulas. - In \(f(2x)\), ask which values of \(x\) make \(2x\) land in each original interval. - In \(f(x-1)\), shift every breakpoint consistently with the input shift.

1. The four graph segments have slopes \(1,-1,1,-1\). Using the visible vertices gives \(f(x)=x+2\) on \([-2,0)\), \(f(x)=-x+2\) on \([0,2)\), \(f(x)=x-2\) on \([2,4)\), and \(f(x)=-x+6\) on \([4,6]\). 2. For \(k(x)=f(2x)-1\), transform each input interval by requiring \(2x\) to lie in the corresponding interval for \(f\). The breakpoints become \(-1,0,1,2,3\). 3. Substituting \(2x\) into each branch and subtracting \(1\) gives \(2x+1\), \(-2x+1\), \(2x-3\), and \(-2x+5\), respectively. Thus, the domain of \(k\) is \([-1,3]\). 4. For \(p(x)=0.5f(x-1)+1\), require \(x-1\) to lie in each original interval. The breakpoints become \(-1,1,3,5,7\). 5. Substituting \(x-1\) into each branch, multiplying by \(0.5\), and adding \(1\) gives \(0.5x+1.5\), \(-0.5x+2.5\), \(0.5x-0.5\), and \(-0.5x+4.5\), respectively. Thus, the domain of \(p\) is \([-1,7]\).

a) \(f(x)=\begin{cases}x+2&-2\le x<0\\-x+2&0\le x<2\\x-2&2\le x<4\\-x+6&4\le x\le6\end{cases}\) b) \(k(x)=\begin{cases}2x+1&-1\le x<0\\-2x+1&0\le x<1\\2x-3&1\le x<2\\-2x+5&2\le x\le3\end{cases}\), domain \([-1,3]\) c) \(p(x)=\begin{cases}0.5x+1.5&-1\le x<1\\-0.5x+2.5&1\le x<3\\0.5x-0.5&3\le x<5\\-0.5x+4.5&5\le x\le7\end{cases}\), domain \([-1,7]\)