Free Math Worksheets

- “Started” means round the number of \(100\,\text{cm}^3\) blocks up. - Determine whether the fee changes continuously or in fixed jumps. - Identify the multiples of \(100\) in the interval.

1. A displacement of \(1200\,\text{cm}^3\) uses exactly \(12\) blocks, so \(H(1200)=12\cdot\$2.00=\$24.00\). 2. Both \(1201\,\text{cm}^3\) and \(1299\,\text{cm}^3\) start a thirteenth block, so each fee is \(13\cdot\$2.00=\$26.00\). 3. The graph is a step function. It stays constant within each \(100\,\text{cm}^3\) block and jumps immediately after each multiple of \(100\). 4. Therefore, the function is discontinuous at \(V=1100, 1200, 1300, 1400\) within the stated interval.

a) \(H(1200)=\$24.00\), \(H(1201)=\$26.00\), and \(H(1299)=\$26.00\) b) \(H\) is a step function and is not continuous. The jump discontinuities are at \(V=1100, 1200, 1300, 1400\).

- What result do you get from direct substitution? - Factor the numerator or denominator to look for a common factor. - If the factor causing the discontinuity cancels completely, what type of discontinuity remains? - If an uncanceled denominator factor approaches zero, compare the behavior from the left and right.

1. For part a, direct substitution gives \(\frac{0}{0}\). Factor: \(f(x)=\frac{(x-3)(x+3)}{2(x-3)}\). 2. For \(x\neq 3\), \(f(x)=\frac{x+3}{2}\). Therefore, \(\lim_{x\to 3}f(x)=\frac{3+3}{2}=3\). The finite limit shows that \(x=3\) is a removable discontinuity. 3. For part b, direct substitution also gives \(\frac{0}{0}\). Factor: \(g(x)=\frac{x-1}{(x-1)^2}\). 4. For \(x\neq 1\), \(g(x)=\frac{1}{x-1}\). As \(x\to 1^-\), \(g(x)\to -\infty\), and as \(x\to 1^+\), \(g(x)\to +\infty\). 5. The two-sided limit does not exist, and \(x=1\) is an infinite discontinuity.

a) \(\lim_{x\to 3}f(x)=3\); the discontinuity is removable. b) \(\lim_{x\to 1}g(x)\) does not exist because the one-sided limits are \(-\infty\) and \(+\infty\); the discontinuity is infinite.

- Check whether the original function has a value at \(x=1\). - Distinguish a limit from an actual function value. - Canceling a factor does not restore an excluded input automatically.

1. The original denominator is \(0\) at \(x=1\), so \(f(1)\) is not defined. 2. Factoring gives \(\frac{(x-1)(x+1)}{x-1}=x+1\) only for \(x\ne1\). 3. Although \(\lim_{x\to1}f(x)=2\), continuity at \(x=1\) requires the function to be defined there and to satisfy \(f(1)=2\). 4. Therefore, the statement is false. The graph has a removable discontinuity, or hole, at \((1, 2)\).

The statement is false. The limit is \(2\), but \(f(1)\) is undefined, so \(f\) is not continuous at \(x=1\). The discontinuity is removable.

- Find the input that makes the denominator \(0\). - Factor and simplify for allowed inputs. - Use the simplified rule to find the missing limiting value.

1. The denominator is \(0\) at \(x=3\), so the domain is \(\mathbb{R}\setminus\{3\}\). 2. Factor the numerator: \(x^2-4x+3=(x-3)(x-1)\). For \(x\ne3\), \(g(x)=x-1\). 3. Therefore, \(\lim_{x\to3}g(x)=\lim_{x\to3}(x-1)=2\). The function values approach \(2\) from both sides. 4. The discontinuity is removable. Defining \(g(3)=2\) makes the extended function continuous.

a) \(\mathbb{R}\setminus\{3\}\) b) The function values approach \(2\) as \(x\to3\). c) Yes. Define \(g(3)=2\).

- Pay attention to whether each boundary value is included in the lower or upper interval. - Use piecewise notation to show a different constant value on each weight interval. - Continuity requires the two one-sided limits to be equal.

1. The package weighing \(1.8\,\text{lb}\) is in the first interval, so its cost is \(\$4.95\). The package weighing \(5.0\,\text{lb}\) is included in the second interval, so its cost is \(\$6.95\). The package weighing \(5.1\,\text{lb}\) is in the third interval, so its cost is \(\$10.45\). 2. The cost function is \(K(w)=\begin{cases}\$4.95&\text{if }0<w\le2\\\$6.95&\text{if }2<w\le5\\\$10.45&\text{if }5<w\le10\\\$18.45&\text{if }10<w\le31.5\end{cases}\). 3. At \(w=5\), \(\lim_{w\to5^-}K(w)=\$6.95\), while \(\lim_{w\to5^+}K(w)=\$10.45\). 4. Since the one-sided limits are different, the two-sided limit does not exist. The function has a jump discontinuity at \(w=5\).

a) \(1.8\,\text{lb}:\ \$4.95\); \(5.0\,\text{lb}:\ \$6.95\); \(5.1\,\text{lb}:\ \$10.45\) b) \(K(w)=\begin{cases}\$4.95&0<w\le2\\\$6.95&2<w\le5\\\$10.45&5<w\le10\\\$18.45&10<w\le31.5\end{cases}\) c) Not continuous, because \(\lim_{w\to5^-}K(w)=\$6.95\ne\$10.45=\lim_{w\to5^+}K(w)\).

- Find the zeros of the denominator to determine the excluded values. - Factor the numerator and denominator before evaluating the limits. - What does a canceled factor imply about the corresponding discontinuity? - Compare the left-hand and right-hand behavior near an uncanceled denominator zero.

1. Factor the denominator: \(x^2-3x+2=(x-1)(x-2)\). Therefore, \(D_f=\mathbb{R}\setminus\{1, 2\}\). 2. Factor the numerator: \(x^2-1=(x-1)(x+1)\). For \(x\neq 1\), \(f(x)=\frac{x+1}{x-2}\). 3. Evaluate the first limit: \(\lim_{x\to 1}f(x)=\frac{1+1}{1-2}=-2\). Because the limit is finite, \(x=1\) is a removable discontinuity. 4. Near \(x=2\), the numerator of the simplified expression approaches \(3\), while the denominator approaches \(0\). Specifically, \(\lim_{x\to 2^-}f(x)=-\infty\) and \(\lim_{x\to 2^+}f(x)=+\infty\). 5. The one-sided limits are different, so \(\lim_{x\to 2}f(x)\) does not exist. The discontinuity at \(x=2\) is infinite.

a) \(D_f=\mathbb{R}\setminus\{1, 2\}\) b) \(\lim_{x\to 1}f(x)=-2\). The limit \(\lim_{x\to 2}f(x)\) does not exist because the one-sided limits are \(-\infty\) and \(+\infty\). c) The discontinuity at \(x=1\) is removable, and the discontinuity at \(x=2\) is infinite.

- Try direct substitution first. - Factor the numerator and denominator when you obtain \(\frac{0}{0}\). - Use one-sided signs when a denominator approaches zero without a canceling factor.

1. Direct substitution gives \(f(0)=\frac{-8}{-16}=\frac{1}{2}\). This is the y-intercept. 2. Factor: \(f(x)=\frac{(x-4)(x+2)}{(x-4)(x+4)}\). For \(x\ne4\), this simplifies to \(\frac{x+2}{x+4}\). Therefore, the limit at \(4\) is \(\frac{6}{8}=\frac{3}{4}\). The original graph has a removable discontinuity, or hole, at \(\left(4, \frac{3}{4}\right)\). 3. Near \(x=-4\), the simplified numerator approaches \(-2\), while the denominator approaches \(0\). From the left, the quotient approaches \(+\infty\); from the right, it approaches \(-\infty\). Therefore, the two-sided limit does not exist, and \(x=-4\) is a vertical asymptote.

1. \(\frac{1}{2}\) 2. \(\frac{3}{4}\); removable discontinuity at \(\left(4, \frac{3}{4}\right)\) 3. The two-sided limit does not exist: the left-hand limit is \(+\infty\), and the right-hand limit is \(-\infty\). The vertical asymptote is \(x=-4\).

- Factor the denominator to find all excluded values. - Factor the numerator and check for common factors. - If the denominator has greater multiplicity than the numerator at a common zero, what remains after simplification? - Use the simplified expression to evaluate the behavior near each excluded value.

1. Factor the denominator: \(x^3-2x^2=x^2(x-2)\). The discontinuities occur at \(x=0\) and \(x=2\). 2. Factor the numerator: \(x^2-4=(x-2)(x+2)\). 3. At \(x=2\), cancel the common factor for \(x\neq 2\): \(g(x)=\frac{x+2}{x^2}\). Then \(\lim_{x\to 2}g(x)=\frac{2+2}{2^2}=1\). Thus, \(x=2\) is a removable discontinuity. 4. At \(x=0\), use the simplified expression \(\frac{x+2}{x^2}\). The numerator approaches \(2\), and \(x^2\to 0^+\) from both sides. Therefore, \(\lim_{x\to 0}g(x)=+\infty\). 5. Thus, \(x=0\) is an infinite discontinuity, the graph does not change sign across it, and the vertical asymptote is \(x=0\).

a) The discontinuities are \(x=0\) and \(x=2\). b) \(\lim_{x\to 2}g(x)=1\), and \(\lim_{x\to 0}g(x)=+\infty\). c) The discontinuity at \(x=2\) is removable. The discontinuity at \(x=0\) is infinite, with vertical asymptote \(x=0\).

- A filled point gives the actual function value; an open circle is excluded. - Look for places where the graph has a hole or a jump. - Compare the values approached from the left and right at each break. - For a removable discontinuity, the limit exists but differs from the function value.

1. At \(x=-2\), the filled point is at \((-2, 3)\), so \(f(-2)=3\). The open circle at \((-2, 1)\) is not included. 2. At \(x=1\), the filled point is at \((1, 0)\), so \(f(1)=0\). The open circle at \((1, 2.5)\) is not included. 3. At \(x=3\), the graph lies on the line segment at \(y=-2\), so \(f(3)=-2\). 4. At \(x=-2\), both one-sided limits equal \(1\), but \(f(-2)=3\). Therefore, \(f\) has a removable discontinuity there. 5. At \(x=1\), the left-hand limit is \(2.5\), while the right-hand limit is \(0\). Therefore, \(f\) has a jump discontinuity there. The graph is continuous at all other inputs in the interval.

a) \(f(-2)=3\), \(f(1)=0\), and \(f(3)=-2\) b) \(f\) is not continuous at \(x=-2\) and \(x=1\). At \(x=-2\), the discontinuity is removable; at \(x=1\), it is a jump discontinuity.

- First determine how the horizontal shift changes the input of the discontinuity. - Reflection across the x-axis changes output values but not input values. - Read the original one-sided limits and function value at \(x=1\). - Multiply the relevant output values by \(-1\).

1. Shifting the graph \(2\) units to the right moves the discontinuity from \(x=1\) to \(x_0=3\). Reflecting across the x-axis does not change its x-coordinate. 2. For the original function, \(\lim_{x\to1^-}f(x)=0\), \(\lim_{x\to1^+}f(x)=1\), and \(f(1)=1\). 3. Reflection across the x-axis multiplies every output by \(-1\). Therefore, \(\lim_{x\to3^-}h(x)=0\), \(\lim_{x\to3^+}h(x)=-1\), and \(h(3)=-1\).

a) \(x_0=3\) b) \(\lim_{x\to3^-}h(x)=0\), \(\lim_{x\to3^+}h(x)=-1\), and \(h(3)=-1\)

- Use filled points to identify actual function values. - Use open circles and nearby graph pieces to identify limiting values. - Compare the left-hand limit, right-hand limit, and function value at each input.

1. At \(x=-1\), the graph approaches \(1\) from both sides, so \(\lim_{x\to-1}g(x)=1\). However, the filled point shows that \(g(-1)=2\). Since the limit and function value are different, \(g\) has a removable discontinuity at \(x=-1\). 2. At \(x=2\), the left-hand limit is \(4\), the right-hand limit is \(0\), and the filled point gives \(g(2)=4\). Since the one-sided limits are different, \(g\) has a jump discontinuity at \(x=2\).

\(g\) is not continuous at either input. It has a removable discontinuity at \(x=-1\) and a jump discontinuity at \(x=2\).

- Reflection across the y-axis changes the sign of each x-coordinate. - Locate the jump in the original graph first. - Reflection across the y-axis reverses the left and right sides of approach. - Compare the two one-sided limits at the reflected input.

1. The graph of \(f\) has a jump at \(x_f=1\): the left-hand limit is \(0\), while the right-hand limit is \(2\). 2. Reflection across the y-axis changes each input \(x\) to \(-x\), so the jump moves from \(x=1\) to \(x_g=-1\). 3. The reflection reverses the sides of approach. Thus \(\lim_{x\to-1^-}g(x)=2\) and \(\lim_{x\to-1^+}g(x)=0\). 4. The filled point on the original graph gives \(f(1)=0\), so \(g(-1)=f(1)=0\). Since the one-sided limits are different, \(g\) is not continuous at \(x=-1\).

a) \(x_f=1\) b) \(x_g=-1\) c) \(\lim_{x\to-1^-}g(x)=2\) and \(\lim_{x\to-1^+}g(x)=0\). Since the one-sided limits are different, \(g\) has a jump discontinuity at \(x=-1\).

- Track how the horizontal translation moves the original jump. - Use the filled point for the function value. - Read the left and right branches separately for the one-sided limits. - Apply the vertical shift to all three y-values.

1. Replacing \(x\) with \(x-3\) shifts the graph right \(3\) units, and adding \(1\) shifts it up \(1\) unit. 2. The jump at \(x=1\) therefore moves to \(x_0=1+3=4\). 3. The filled point on the original graph is \((1, -5)\), so \(k(4)=f(1)+1=-5+1=-4\). 4. The left-hand limit of \(f\) at \(1\) is \(-5\), so \(\lim_{x\to4^-}k(x)=-5+1=-4\). 5. The right-hand limit of \(f\) at \(1\) is \(-1\), so \(\lim_{x\to4^+}k(x)=-1+1=0\).

a) Shift right \(3\) units and up \(1\) unit. b) \(x_0=4\) c) \(k(4)=-4\), \(\lim_{x\to4^-}k(x)=-4\), and \(\lim_{x\to4^+}k(x)=0\)

- An open circle shows a value the graph approaches but does not include. - A filled point gives the actual function value. - Continuity requires the limit to equal the function value.

1. As \(x\) approaches \(2\) from either side, the graph approaches \(3\). Therefore, \(\lim_{x\to2}g(x)=3\). 2. The filled point gives the actual function value \(g(2)=1\). 3. Since \(\lim_{x\to2}g(x)\ne g(2)\), the function has a removable discontinuity at \(x=2\). 4. To make the function continuous, the missing point on the line must be filled, so the function must be redefined with \(g(2)=3\).

a) \(g\) is not continuous at \(x=2\) because \(\lim_{x\to2}g(x)=3\), but \(g(2)=1\). b) \(g(2)=3\)