- Positive function values appear where the graph is above the x-axis. - Use both the x-intercept and the vertical asymptote as interval boundaries. - Check whether either boundary can be included.

1. The graph crosses the x-axis at \(x=-1\) and has a vertical asymptote at \(x=2\). 2. The graph is above the x-axis for \(x<-1\) and for \(x>2\). 3. The zero at \(x=-1\) is not included because the inequality is strict, and \(x=2\) is not in the domain.

\((-\infty,-1)\cup(2,\infty)\)

- Sign-chart boundaries come from zeros of the numerator and zeros of the denominator. - A denominator zero is never part of the rational function's domain.

1. The numerator is zero at \(x=3\). 2. The denominator is zero at \(x=-1\). 3. Both values must be marked on the sign chart, but \(x=-1\) is excluded from the domain.

Mark \(x=-1\) and \(x=3\). The excluded value is \(x=-1\).

- What part of a fraction must not equal zero? - What value does the whole fraction have at the proposed input? - Does the inequality allow equality?

1. At \(x=4\), the denominator is \(4+3=7\), so the fraction is defined. 2. Substituting gives \(\frac{4-4}{4+3}=\frac{0}{7}=0\). 3. Since \(0\ge0\), the value \(x=4\) satisfies the inequality. The student's claim is incorrect.

No. The fraction equals \(0\) at \(x=4\), so \(x=4\) satisfies the inequality.

- What happens when the denominator of a fraction is zero? - Set only the denominator equal to zero to find the forbidden input. - Can an undefined expression be positive?

1. The denominator is zero when \(x-5=0\), so \(x=5\) is excluded. 2. At this input the fraction is undefined, so it cannot satisfy the inequality, no matter what the inequality sign is.

\(x=5\) is excluded and cannot be in the solution set.

- What input separates the fraction's domain into two intervals? - Evaluate the expression at the test input from each interval. - Which interval gives a negative value? Check whether the boundary itself is allowed.

1. The denominator is zero at \(x=2\), so the two intervals to check are \((-\infty, 2)\) and \((2, \infty)\). 2. At \(x=0\), the fraction is \(\frac{6}{-2}=-3<0\). At \(x=4\), it is \(\frac{6}{2}=3>0\). 3. The fraction is negative only for \(x<2\); the undefined value \(x=2\) is excluded.

\((-\infty, 2)\)

- Check the graph's height on either side of the point where it touches the x-axis. - Does touching the axis always mean the function changes sign? - Does a strict inequality include a point where the output is zero?

1. The graph touches the x-axis at \((1,0)\) and remains above it on either side. 2. Therefore, the function is positive at every real input except \(x=1\), where its value is zero.

The graph touches at \(x=1\). The solution is \((-\infty,1)\cup(1,\infty)\).

- Which inputs make the numerator zero, and what output do they give? - Check whether the expression is defined at each input. - Compare what \(>\) and \(\ge\) allow when the output is zero.

1. At \(x=-3\) and \(x=2\), the numerator is zero while the denominator is not. Therefore, \(r(-3)=r(2)=0\). 2. At \(x=1\), the denominator is zero, so \(r(1)\) is undefined. 3. Both zeros satisfy the non-strict inequality, but none of the three values satisfy the strict inequality.

a) \(x=-3\) and \(x=2\) b) None of the three values

- Check the denominator in the original fraction, before canceling. - Which marked value makes that denominator zero? - Does simplifying an expression change where the original fraction was undefined?

1. The original denominator is zero at \(x=-2\), which is marked \(A\). The original fraction is undefined there. 2. Canceling the common factor gives \(x-3\) only for allowed inputs, so the exclusion at \(A\) remains.

Point \(A\), at \(x=-2\); the original denominator is zero there.

- Which input makes the denominator zero? - Substitute each requested input into the original expression. - For which inputs is the denominator negative?

1. At \(x=3\), the denominator is zero, so this value is excluded from the domain. 2. \(T(0)=-5\), \(T(4)=15\), and \(T(8)=3\). 3. The numerator is positive. The fraction is negative exactly when \(x-3<0\), which means \(x<3\).

a) \(x=3\) makes the denominator zero. b) \(T(0)=-5\), \(T(4)=15\), \(T(8)=3\). c) \(x\in(-\infty,3)\). Since \(15\) is positive, the fraction is negative exactly when \(x-3<0\).

- How can you clear a denominator when solving for \(x\)? - When multiplying an inequality by \(x\), why does the condition \(x > 0\) matter? - What happens to \(\frac{10}{x}\) as positive \(x\) increases from \(10\) to \(100\)?

1. Solve \(-\frac{10}{x} + 2 = 7\). Then \(-\frac{10}{x} = 5\), so \(-10 = 5x\) and \(x = -2\). 2. For \(x > 0\), solve \(-\frac{10}{x} + 2 > 0\). This is equivalent to \(2 > \frac{10}{x}\). Multiplying by the positive quantity \(x\) gives \(2x > 10\), so \(x > 5\). 3. For positive \(x\), the quantity \(\frac{10}{x}\) decreases as \(x\) increases. Since \(g(x) = 2 - \frac{10}{x}\), subtracting the smaller quantity gives the larger output. Therefore, \(g(100) > g(10)\).

a) \(x = -2\) b) \(x > 5\) c) \(g(100) > g(10)\), because \(\frac{10}{x}\) decreases as positive \(x\) increases.

- Start by identifying the value that makes the denominator zero. - For a quotient with a negative numerator, what sign must the denominator have to make the quotient positive? - Use the denominator's sign on each side of the excluded value to determine the sign of the quotient.

1. The expression is undefined when the denominator is zero: \(x+4=0\), so \(x=-4\). 2. Because the numerator is negative, the quotient is positive when the denominator is negative. Solve \(x+4<0\) to get \(x<-4\), so the solution is \((-\infty,-4)\). 3. The quotient is negative when the denominator is positive. Solve \(x+4>0\) to get \(x>-4\), so the solution is \((-4,\infty)\).

1. \(\{-4\}\) 2. \((-\infty,-4)\) 3. \((-4,\infty)\)

- First find the x-value that makes the denominator zero. - With a positive numerator, which sign of the denominator makes the quotient negative? - Compare the new expression with the original at one allowed input. What stays the same, and what changes?

1. The denominator is zero when \(2x-10=0\), so \(x=5\). Therefore, the domain is \(\mathbb{R}\setminus\{5\}\). 2. Because the numerator \(8\) is positive, the quotient is negative when the denominator is negative. Solve \(2x-10<0\) to get \(x<5\), so the solution is \((-\infty,5)\). 3. Replacing \(8\) by \(-8\) multiplies every defined output by \(-1\), so the signs reverse. The new expression is positive on \((-\infty,5)\) and negative on \((5,\infty)\). The excluded value \(x=5\) does not change.

a) \(\mathbb{R}\setminus\{5\}\) b) \((-\infty,5)\) c) The signs reverse: positive on \((-\infty,5)\) and negative on \((5,\infty)\).

- What is the smallest possible value of \(x^2+1\)? - If the denominator is always positive, which factor determines the sign of the quotient? - For the modified numerator, decide whether \(x^2+5\) can ever equal zero for a real x-value.

1. For every real \(x\), \(x^2\ge0\), so \(x^2+1\ge1\). The denominator is never zero, so the domain is \(\mathbb{R}\). 2. Because the denominator is always positive, the sign of \(Q(x)\) is determined by \(x-5\). Thus \(Q(x)<0\) for \(x<5\), \(Q(x)=0\) at \(x=5\), and \(Q(x)>0\) for \(x>5\). 3. If the numerator were \(x^2+5\), then it would be at least \(5\) for every real \(x\), so it could never be zero.

a) The domain is \(\mathbb{R}\). b) Negative on \((-\infty,5)\); positive on \((5,\infty)\); zero at \(x=5\). c) \(x^2+5\) is always positive, so the modified expression is never zero.

- A rational expression is zero when its numerator is zero and its denominator is nonzero. - Decide what sign \(x^2+1\) has for every real x-value. - In part (b), compare the sign of the numerator on the interval \(x>3\).

1. The denominator \(x^2+1\) is positive for every real \(x\), so \(T(x)=0\) exactly when the numerator is zero. Solve \(4x-12=0\) to get \(x=3\). 2. If \(x>3\), then \(4x-12>0\). The denominator is also positive, so \(T(x)>0\). 3. Substitute \(x=-1\): \(T(-1)=\frac{4(-1)-12}{(-1)^2+1}=\frac{-16}{2}=-8\).

a) \(x=3\) b) For \(x>3\), both the numerator and denominator are positive, so \(T(x)>0\). c) \(T(-1)=-8\)

- Record all restrictions from the original denominator before simplifying. - Cancellation can remove a factor from the sign calculation without restoring the excluded input to the domain. - Use the remaining zero and denominator zero to test the signs of the intervals.

1. The original denominator excludes \(x=3\) and \(x=2\). 2. For sign analysis, cancel the common factor to obtain \(\frac{x+3}{x-2}\), while keeping \(x=3\) excluded. 3. The remaining zero is \(x=-3\), and the vertical asymptote is \(x=2\). 4. The simplified expression is nonnegative on \((-\infty,-3]\) and \((2,\infty)\). 5. Remove the hole at \(x=3\) from the second interval.

\((-\infty,-3]\cup(2,3)\cup(3,\infty)\)

- Determine the sign of the squared denominator on either side of its zero. - Decide where the numerator is negative or zero. - Keep the denominator zero excluded even if the surrounding interval satisfies the inequality.

1. The numerator is zero at \(x=3\), which may be included because equality is allowed. 2. The denominator is zero at \(x=1\), so \(x=1\) is excluded. 3. Because \((x-1)^2>0\) for every \(x\ne1\), the denominator does not change the sign of the expression. 4. The expression is nonpositive where \(x-3\le0\), except at the excluded value \(x=1\).

\((-\infty,1)\cup(1,3]\)

- Find the input where the numerator is zero and the input where the denominator is zero. - Pick a test number from each interval those values create. - Check which boundary can be included with the symbol \(\le\).

1. The numerator is zero at \(x=1\), and the denominator is zero at \(x=-2\). 2. The expression is positive on \((-\infty, -2)\), negative on \((-2, 1)\), and positive on \((1, \infty)\). 3. Include \(x=1\) because the fraction equals zero there. Exclude \(x=-2\) because it is undefined. The solution is \((-2, 1]\).

Signs from left to right: positive, negative, positive. Solution: \((-2, 1]\).

- Compare the signs of the numerators at the two test values. - What changes when a factor is squared? - Is the denominator positive or negative at those test values?

1. At \(x=0\), \(r(0)=-\frac{1}{2}<0\) and \(s(0)=\frac{1}{2}>0\). 2. At \(x=2\), \(r(2)=\frac{1}{4}>0\) and \(s(2)=\frac{1}{4}>0\). 3. The factor \(x-1\) has odd multiplicity in \(r\), so its sign changes at \(1\). The squared factor in \(s\) does not change sign at \(1\); both sides are positive near that zero.

a) At \(x=0\): \(r<0\), \(s>0\). At \(x=2\): both are positive. b) Only \(r\) changes sign at \(x=1\).

- Compare what the curve does on both sides of each zero. - Look for an input where separate branches do not meet. - Check every interval for positive outputs, then decide which boundary inputs are allowed.

1. The zeros are \(x=-1\) and \(x=3\); the vertical asymptote is \(x=1\). At \(x=-1\), the graph touches the axis and stays above it on both sides. 2. The graph is positive on \((-\infty,-1)\), \((-1,1)\), and \((3,\infty)\), but negative on \((1,3)\). 3. Exclude both zeros because the inequality is strict. Exclude \(x=1\) because the function is undefined there.

a) Zeros: \(x=-1\) (touches, no sign change) and \(x=3\) (crosses). Vertical asymptote: \(x=1\). b) \((-\infty,-1)\cup(-1,1)\cup(3,\infty)\).

- Identify the zero of the numerator and the zero of the denominator separately. - Test an input from each interval they create. - Which endpoint is a valid zero, and which one makes the fraction undefined?

1. The numerator is zero at \(x=1\); the denominator is zero at \(x=4\). 2. At \(x=0\), the fraction is positive; at \(x=2\), it is negative; at \(x=5\), it is positive. 3. Therefore, the fraction is nonnegative for \(x\le1\) or \(x>4\). Maya incorrectly included \(x=4\), where the denominator is zero.

\((-\infty, 1]\cup(4, \infty)\). Maya must exclude \(x=4\) because the fraction is undefined there.

- Factor the numerator and denominator completely. - The sign can change only at a zero or an excluded value. - Test one value in each interval determined by the critical values. - Include numerator zeros, but exclude denominator zeros.

1. Factor the numerator: \(x^3-4x=x(x^2-4)=x(x-2)(x+2)\). The numerator is zero at \(x=-2\), \(x=0\), and \(x=2\). 2. Factor the denominator: \(x^2-2x-3=(x-3)(x+1)\). The excluded values are \(x=-1\) and \(x=3\). No factors cancel, so the x-intercepts are \((-2, 0)\), \((0, 0)\), and \((2, 0)\). 3. Use the critical values \(-2\), \(-1\), \(0\), \(2\), and \(3\) to make a sign chart for \(\frac{x(x-2)(x+2)}{(x-3)(x+1)}\). 4. The expression is nonnegative on \([-2, -1)\), \([0, 2]\), and \((3, \infty)\). Include the zeros of the numerator and exclude the zeros of the denominator.

a) The x-intercepts are \((-2, 0)\), \((0, 0)\), and \((2, 0)\). b) \([-2, -1)\cup[0, 2]\cup(3, \infty)\)

- Move all terms to one side and combine them into one rational expression. - Identify both numerator zeros and denominator zeros before making a sign chart. - Test one value in each interval determined by the critical values. - Include a numerator zero when equality is allowed, but never include a value where the original denominator is zero.

1. Move all terms to one side: \(\frac{3}{x-2}-x\ge0\). 2. Combine the terms: \(\frac{3-x(x-2)}{x-2}=\frac{-(x-3)(x+1)}{x-2}\ge0\). 3. The critical values are \(x=-1\) and \(x=3\), where the numerator is zero, and \(x=2\), where the denominator is zero. 4. Test the intervals determined by \(-1\), \(2\), and \(3\). The expression is nonnegative on \((-\infty,-1]\) and \((2,3]\). 5. The value \(x=2\) is excluded because the original rational expression is undefined there.

The solution is \((-\infty,-1]\cup(2,3]\). The value \(x=2\) is excluded because it makes the denominator zero.

- Factor the numerator. - Find the value that makes the denominator zero. - Use the zeros and excluded value to create test intervals. - Decide which boundary values can be included.

1. Factor the numerator: \(x^2-2x-8=(x-4)(x+2)\). Its zeros are \(x=-2\) and \(x=4\). 2. The denominator is zero at \(x=1\), so \(x=1\) is excluded. 3. The critical values \(-2\), \(1\), and \(4\) divide the number line into test intervals. 4. A sign chart for \(\frac{(x-4)(x+2)}{x-1}\) gives a negative sign on \((-\infty, -2)\), a positive sign on \((-2, 1)\), a negative sign on \((1, 4)\), and a positive sign on \((4, \infty)\). 5. Include the numerator zeros because equality is allowed, and exclude \(x=1\). Therefore, the solution is \([-2, 1)\cup[4, \infty)\).

\([-2, 1)\cup[4, \infty)\)

- Factor both polynomials. - Determine whether the squared numerator can ever be negative. - Find where the denominator is negative. - Exclude values that make the denominator zero.

1. Factor the numerator and denominator: \(x^2-4x+4=(x-2)^2\) and \(x^2-x-6=(x-3)(x+2)\). 2. The numerator is nonnegative for every real \(x\) and equals \(0\) only at \(x=2\). The excluded values are \(x=-2\) and \(x=3\). 3. The denominator \((x-3)(x+2)\) is negative between its zeros, so it is negative on \((-2, 3)\). 4. On \((-2, 3)\), the quotient is negative except at \(x=2\), where it equals \(0\). Thus every value in that open interval satisfies the inequality. 5. The endpoints are excluded because they make the denominator zero. Therefore, the solution is \((-2, 3)\).

\((-2, 3)\)

- Write the domain restrictions from the original expression before canceling anything. - Ask whether canceling a common factor changes the original domain. - After solving the simplified inequality, remove every value excluded from the original expression.

1. The original denominator is zero at \(x=0\) and \(x=1\), so both values are excluded from the domain. 2. Canceling \(x\) is valid for sign analysis only when the original restriction \(x\ne0\) is retained. 3. The simplified expression \(\frac{x+2}{x-1}\) is negative on \((-2,1)\). 4. Removing the excluded value \(x=0\) gives \((-2,0)\cup(0,1)\).

Jordan forgot that \(x=0\) is excluded by the original denominator. The correct solution is \((-2,0)\cup(0,1)\).

- How does the current change the boat's speed upstream and downstream? - Write each travel time as distance divided by speed. - Before multiplying an inequality by expressions involving \(v\), check whether those expressions are positive.

1. The upstream speed is \(v-3\) mph and the downstream speed is \(v+3\) mph. Their travel times are \(\frac{12}{v-3}\) and \(\frac{12}{v+3}\) hours. 2. The condition is \(\frac{12}{v-3}\le\frac{24}{v+3}\). 3. Since \(v>3\), both denominators are positive. Multiply through to get \(12(v+3)\le24(v-3)\), then \(v+3\le2v-6\), so \(v\ge9\). 4. This also meets the original restriction \(v>3\).

\(v\ge9\,\text{mph}\)

- Which factors would put zeros at the required inputs? - Which factors would make the denominator zero at the required vertical asymptotes? - Test a point in one required positive interval. Does your first proposed rule have the correct sign there?

1. Zeros at \(-3\) and \(4\) suggest numerator factors \((x+3)(x-4)\). Poles at \(-1\) and \(1\) suggest denominator factors \((x+1)(x-1)\). 2. The ratio \(\frac{(x+3)(x-4)}{(x+1)(x-1)}\) is negative between \(-3\) and \(-1\) and between \(1\) and \(4\). Negating it makes it positive on those two intervals. 3. Thus, \(r(x)=-\frac{(x+3)(x-4)}{(x+1)(x-1)}\). Its sign is negative, positive, negative, positive, negative on the five intervals separated by the four critical inputs. 4. Include the two zeros and exclude the two denominator zeros. The nonnegative solution set is exactly \([-3,-1)\cup(1,4]\).

One possible function is \(r(x)=-\frac{(x+3)(x-4)}{(x+1)(x-1)}\).

- Compare the fraction with \(1\) by bringing both sides to a common denominator. - For \(x>2\), what is the sign of the denominator? - After finding the allowed parameters, check separately whether the resulting numerator can equal zero. - Keep the original excluded input in every case.

1. The fraction is undefined at \(x=2\). For all other inputs, subtracting \(1\) gives \(\frac{x-k}{x-2}-1=\frac{2-k}{x-2}\). 2. When \(x>2\), the denominator is positive. Therefore, the inequality holds for every such input exactly when \(2-k\ge0\), or \(k\le2\). 3. If \(k<2\), the numerator \(2-k\) is positive, so \(\frac{2-k}{x-2}\ge0\) exactly for \(x>2\). 4. If \(k=2\), the numerator is zero. The original fraction equals \(1\) for every \(x\ne2\), so all those inputs satisfy the inequality.

\(k\le2\). If \(k<2\), the solution set is \((2,\infty)\). If \(k=2\), the solution set is \((-\infty,2)\cup(2,\infty)\).