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Find each \( x \)-value at which \( f \) is discontinuous and for each \( x \)-value, determine whether \( f \) is continuous from the right, or from the left, or neither. \[ f(x)=\left\{\begin{array}{ll} x+2 & \text { if } x<1 \\ 1 / x & \text { if } 1
Ask by Ruiz May. in the United States
Feb 03,2025

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- **Discontinuities at:** - \( x = 1 \): Neither continuous from the right nor from the left. - \( x = 2 \): Continuous from the right only. - **Graph Sketch:** - For \( x < 1 \): Line \( f(x) = x + 2 \) with an open circle at \( (1, 3) \). - For \( 1 < x < 2 \): Hyperbola \( f(x) = \frac{1}{x} \) with an open circle at \( (1, 1) \) and approaching \( (2, 0.5) \). - For \( x \geq 2 \): Square root function \( f(x) = \sqrt{x - 2} \) starting at \( (2, 0) \) with a closed circle.

Solución

To analyze the continuity of the function \( f(x) \) and sketch its graph, let's examine each piece and identify potential points of discontinuity. ### Given Function: \[ f(x)=\left\{ \begin{array}{ll} x+2 & \text{if } x < 1 \\ \frac{1}{x} & \text{if } 1 < x < 2 \\ \sqrt{x-2} & \text{if } x \geq 2 \end{array} \right. \] ### Points of Potential Discontinuity: The potential points of discontinuity are at the boundaries where the pieces of the function meet, namely at \( x = 1 \) and \( x = 2 \). --- ### 1. **At \( x = 1 \) (Smaller Value):** - **Left-Hand Limit (as \( x \to 1^- \)):** \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 2) = 1 + 2 = 3 \] - **Right-Hand Limit (as \( x \to 1^+ \)):** \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \left( \frac{1}{x} \right) = \frac{1}{1} = 1 \] - **Function Value at \( x = 1 \):** The function \( f(x) \) is **not defined** at \( x = 1 \) based on the given piecewise definition. - **Conclusion for \( x = 1 \):** - **Continuous from the Right:** **No** - **Continuous from the Left:** **No** - **Result:** **Neither** --- ### 2. **At \( x = 2 \) (Larger Value):** - **Left-Hand Limit (as \( x \to 2^- \)):** \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \left( \frac{1}{x} \right) = \frac{1}{2} = 0.5 \] - **Right-Hand Limit (as \( x \to 2^+ \)):** \[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \sqrt{x - 2} = \sqrt{2 - 2} = 0 \] - **Function Value at \( x = 2 \):** \[ f(2) = \sqrt{2 - 2} = 0 \] - **Conclusion for \( x = 2 \):** - **Continuous from the Right:** **Yes** (since \( \lim_{x \to 2^+} f(x) = f(2) \)) - **Continuous from the Left:** **No** (since \( \lim_{x \to 2^-} f(x) \neq f(2) \)) - **Result:** **Continuous from the Right** --- ### Summary of Discontinuities: | \( x \)-value | Continuous from the Right | Continuous from the Left | Neither | |---------------|---------------------------|--------------------------|---------| | \( x = 1 \) | No | No | **Yes** | | \( x = 2 \) | **Yes** | No | No | --- ### Graph of \( f(x) \): 1. **For \( x < 1 \): \( f(x) = x + 2 \)** - This is a straight line with a slope of 1 and a y-intercept at \( (0, 2) \). - As \( x \to 1^- \), \( f(x) \to 3 \). - **Open Circle at \( x = 1 \):** Since \( f(1) \) is not defined, represent with an open circle at \( (1, 3) \). 2. **For \( 1 < x < 2 \): \( f(x) = \frac{1}{x} \)** - This is a hyperbola segment decreasing from \( x = 1 \) to \( x = 2 \). - At \( x = 1 \), \( f(x) = 1 \) (but not included, so use an open circle at \( (1, 1) \)). - At \( x = 2 \), \( f(x) = 0.5 \). 3. **For \( x \geq 2 \): \( f(x) = \sqrt{x - 2} \)** - This is the right half of a parabola starting at \( (2, 0) \). - At \( x = 2 \), \( f(x) = 0 \) (use a closed circle at \( (2, 0) \)). **Visual Representation:** ``` y | 4 + | 3 + ● ← Open circle at (1,3) | / 2 +------+---------+---------+ | / \ 1 + ● √(x-2) | 0 +---------------------------- x 1 2 ``` - **Open circles** indicate points that are not included in the function. - **Closed circles** indicate points that are included in the function. - The dashed lines represent the behavior of the function approaching the discontinuities. --- ### Final Answer: - **Discontinuities at:** - \( \boxed{x=1} \) - **Neither continuous from the right nor from the left.** - \( \boxed{x=2} \) - **Continuous from the right only.** - **Graph Sketch:** - **For \( x < 1 \):** Line \( f(x) = x + 2 \) with an open circle at \( (1, 3) \). - **For \( 1 < x < 2 \):** Hyperbola \( f(x) = \frac{1}{x} \) with an open circle at \( (1, 1) \) and approaching \( (2, 0.5) \). - **For \( x \geq 2 \):** Square root function \( f(x) = \sqrt{x - 2} \) starting at \( (2, 0) \) with a closed circle.

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Bonus Knowledge

To determine where the function \( f \) is discontinuous, we need to examine the points where the definition of the function changes, which occur at \( x = 1 \) and \( x = 2 \). 1. At \( x = 1 \): - As \( x \) approaches \( 1 \) from the left (\( x < 1 \)), \( f(x) = x + 2 \), so \( f(1^-) = 1 + 2 = 3 \). - As \( x \) approaches \( 1 \) from the right (\( 1 < x < 2 \)), \( f(x) = \frac{1}{x} \), so \( f(1^+) = \frac{1}{1} = 1 \). - Since \( f(1^-) \) and \( f(1^+) \) do not match, \( f \) is not continuous at \( x = 1 \). Specifically, it's continuous from the left but not from the right. 2. At \( x = 2 \): - As \( x \) approaches \( 2 \) from the left (\( 1 < x < 2 \)), \( f(x) = \frac{1}{x} \), so \( f(2^-) = \frac{1}{2} = 0.5 \). - As \( x \) approaches \( 2 \) from the right (\( x \geq 2 \)), \( f(x) = \sqrt{x - 2} \), so \( f(2^+) = \sqrt{2 - 2} = \sqrt{0} = 0 \). - Here, \( f(2^-) \) and \( f(2^+) \) also do not match, indicating \( f \) is discontinuous at \( x = 2 \). It is continuous from the right but not from the left. So, the discontinuous points are: - \( x = 1 \): continuous from the left, not continuous from the right. - \( x = 2 \): not continuous from the left, continuous from the right. ### Summary: - \( x = 1 \): continuous from left, not from right. - \( x = 2 \): not continuous from left, continuous from right. Now to sketch the graph of \( f(x) \): - When \( x < 1 \), the graph is a straight line that approaches the point (1, 3). - For \( 1 < x < 2 \), the graph is a hyperbola (portion of \( 1/x \)) approaching (2, 0.5). - At \( x = 2 \), there is a jump to (2, 0) since \( \sqrt{x-2} \) starts at (2, 0) for \( x \geq 2). \[ \begin{align*} f(x) & = \text{The graph is a line for } x < 1 \text{ (up to (1, 3)).} \\ & = \text{Curve of } \frac{1}{x} \text{ for } 1 < x < 2 \text{ (down to (2, 0.5)).} \\ & = \text{Then a curve starting at } (2, 0) \text{ that goes rightward.} \end{align*} \] Make sure to indicate jumps at \( x=1 \) and \( x=2 \) as part of your graph sketch!

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