Free Math Worksheets

- 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.

- 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. The 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 \(4\)

- Determine the possible values of \(2+\sin(x)\). - The input of a logarithm must be positive. - Use the fact that \(\ln(x)\) is increasing.

1. The logarithm requires \(2+\sin(x)>0\). Since \(-1\le\sin(x)\le1\), the input satisfies \(1\le2+\sin(x)\le3\), so it is positive for every real \(x\). Thus the domain is \(\mathbb{R}\). 2. The natural logarithm is increasing. Applying it to the interval \([1, 3]\) gives outputs from \(\ln(1)=0\) through \(\ln(3)\). Therefore, the range is \([0, \ln(3)]\).

Domain: \(\mathbb{R}\) Range: \([0, \ln(3)]\)

- 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 \(D_g=[-2, -1)\cup(-1, \infty)\).

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

- 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: \(z(x)=h(f(x))=x^2-1\)

- 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 \((-1, 3)\)

- 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) \(u_4(x)=h(g(x))\)

- 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, \(D_g=(-\infty, -1)\cup(1, \infty)\).

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

- Use a point on the red graph to determine the coefficient in \(h(x)=ax^3\). - Compare \(h(x)\) with \(af(x)\). - Rewrite the coefficient \(8\) as a cube to express \(h\) in the form \(f(cx)\).

1. The red graph passes through \((1, 8)\). If \(h(x)=ax^3\), then \(8=a(1)^3\), so \(h(x)=8x^3\). 2. Since \(h(x)=8f(x)\), it is a vertical stretch by a factor of \(8\). 3. Also, \(h(x)=f(2x)=(2x)^3=8x^3\). Replacing \(x\) with \(2x\) produces a horizontal compression by a factor of \(\frac{1}{2}\).

\(h(x)=8x^3\) Vertical stretch factor: \(8\) Horizontal compression factor: \(\frac{1}{2}\)

- Changes inside the function affect x-coordinates; changes outside affect y-coordinates. - Determine how each endpoint of the domain moves. - Transform the vertices one coordinate at a time.

1. The original graph has vertices \((-2, 0)\), \((0, 2)\), \((2, 0)\), \((4, 2)\), and \((6, 0)\), so its range is \([0, 2]\). 2. For \(k(x)=f(2x)-1\), halve each x-coordinate and subtract \(1\) from each y-coordinate. The new vertices are \((-1, -1)\), \((0, 1)\), \((1, -1)\), \((2, 1)\), and \((3, -1)\). Thus, the domain is \([-1, 3]\) and the range is \([-1, 1]\). 3. For \(p(x)=0.5f(x-1)+1\), shift right \(1\), vertically compress by \(0.5\), and shift up \(1\). The vertices become \((-1, 1)\), \((1, 2)\), \((3, 1)\), \((5, 2)\), and \((7, 1)\). Thus, the domain is \([-1, 7]\) and the range is \([1, 2]\).

a) Domain: \([-1, 3]\); range: \([-1, 1]\); vertices \((-1, -1)\), \((0, 1)\), \((1, -1)\), \((2, 1)\), \((3, -1)\) b) Domain: \([-1, 7]\); range: \([1, 2]\); vertices \((-1, 1)\), \((1, 2)\), \((3, 1)\), \((5, 2)\), \((7, 1)\)