Free math worksheets

- 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 \(x=-1\).

- 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? - Changing only the sign of the numerator changes the sign of every defined output but not the domain.

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 \(x=1\).

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

- 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 \((-2, 3)\).

\((-2, 3)\)

- Use the zero and asymptotes to construct \(k\). - The input of a natural logarithm must be positive. - Make a sign chart using the numerator zero and denominator zero.

1. The graph has vertical asymptote \(x=2\), zero \(x=5\), and horizontal asymptote \(y=1\). A matching rule is \(k(x)=\frac{x-5}{x-2}\). It also gives \(k(0)=\frac{-5}{-2}=2.5\), matching the graph. 2. The logarithm requires \(k(x)>0\), so solve \(\frac{x-5}{x-2}>0\). 3. The critical values are \(2\) and \(5\). The numerator and denominator have the same sign when \(x<2\) or \(x>5\). 4. Therefore, the domain is \((-\infty, 2)\cup(5, \infty)\), which is exactly \(\mathbb{R}\setminus[2, 5]\).

One possible rule is \(k(x)=\frac{x-5}{x-2}\). The domain of \(g\) is \((-\infty, 2)\cup(5, \infty)=\mathbb{R}\setminus[2, 5]\).

- The logarithm is defined where its input is positive. - Identify the graph that lies above the x-axis only between \(-4\) and \(2\). - Use the zero and vertical asymptote to construct the rational function. - Substitute a clear point to determine the scale factor.

1. The logarithm is defined exactly where \(k(x)>0\). Therefore, the correct graph must lie above the x-axis only for \(-4<x<2\). 2. Graph A has a zero at \(x=-4\), a vertical asymptote at \(x=2\), and positive values only between these two values. Graph B is positive outside that interval. Therefore, Graph A is correct. 3. A rule with zero \(-4\) and vertical asymptote \(2\) is \(k(x)=a\frac{x+4}{2-x}\). 4. Graph A passes through \((0, 2)\), so \(2=a\frac{4}{2}=2a\). Thus, \(a=1\), and \(k(x)=\frac{x+4}{2-x}\).

Graph A; \(k(x)=\frac{x+4}{2-x}\)

- 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 \((-2,0)\cup(0,1)\).