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Using the definition of continuity to explain why or why not the following is continuous \( a=3 \). If the function is discontinuous state the type of discontinuity. Show work in WRITTEN WORK! \[ h(x)=\left\{\begin{array}{ll}x^{3}+2 x-33 & \text { if } x \leq 3 \\ \frac{x^{2}-6 x+9}{x-3} & \text { if } x>3\end{array}\right. \]

Ask by Riley Hall. in the United States
Mar 19,2025

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The function \( h(x) \) is continuous at \( x = 3 \) because the left-hand and right-hand limits both equal \( h(3) = 0 \).

Solution

**Step 1. Evaluate \( h(3) \):** Since the piecewise function is defined as \[ h(x)= \begin{cases} x^3+2x-33 & \text{if } x \leq 3,\\[1mm] \frac{x^2-6x+9}{x-3} & \text{if } x>3, \end{cases} \] for \( x = 3 \) we use the first formula: \[ h(3) = 3^3 + 2(3) - 33. \] Calculating, \[ 3^3 = 27,\quad 2(3) = 6,\quad \text{and } 27 + 6 - 33 = 0. \] Thus, \[ h(3)=0. \] **Step 2. Compute the Left-Hand Limit as \( x \to 3^- \):** For \( x < 3 \), we use \( h(x) = x^3+2x-33 \). Since this is a polynomial (which is continuous everywhere), we have: \[ \lim_{x \to 3^-} h(x) = 3^3 + 2(3) - 33 = 27 + 6 - 33 = 0. \] **Step 3. Compute the Right-Hand Limit as \( x \to 3^+ \):** For \( x > 3 \), the function is given by: \[ h(x) = \frac{x^2-6x+9}{x-3}. \] Factor the numerator: \[ x^2-6x+9 = (x-3)^2. \] Thus, for \( x>3 \) (where \( x-3\neq 0 \)): \[ h(x)=\frac{(x-3)^2}{x-3} = x-3. \] Now, take the limit: \[ \lim_{x \to 3^+} h(x) = \lim_{x \to 3^+} (x-3)= 3-3= 0. \] **Step 4. Apply the Definition of Continuity at \( x=3 \):** A function is continuous at \( x=a \) if: 1. \( h(a) \) exists. 2. \(\lim_{x \to a} h(x)\) exists. 3. \(\lim_{x \to a} h(x) = h(a)\). Here: - \( h(3) = 0 \). - \(\lim_{x \to 3^-} h(x) = 0\) and \(\lim_{x \to 3^+} h(x) = 0\) so \(\lim_{x \to 3} h(x)= 0\). - Thus, \(\lim_{x \to 3} h(x) = h(3) = 0\). **Conclusion:** The function \( h(x) \) is continuous at \( x = 3 \) since all conditions of continuity are satisfied.

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The Deep Dive

To determine if the function \( h(x) \) is continuous at \( a = 3 \), we can use the definition of continuity, which states that a function \( h \) is continuous at a point \( c \) if: 1. \( h(c) \) is defined. 2. The limit of \( h(x) \) as \( x \) approaches \( c \) exists. 3. The limit of \( h(x) \) as \( x \) approaches \( c \) equals \( h(c) \). Let's evaluate this for \( a = 3 \). 1. **Calculate \( h(3) \)** using the first piece of the function (since \( x \leq 3 \)): \[ h(3) = 3^3 + 2(3) - 33 = 27 + 6 - 33 = 0. \] 2. **Calculate the limit as \( x \) approaches 3 from the left (\( x \to 3^- \))**: \[ \lim_{x \to 3^-} h(x) = \lim_{x \to 3} (x^3 + 2x - 33) = 3^3 + 2(3) - 33 = 27 + 6 - 33 = 0. \] 3. **Calculate the limit as \( x \) approaches 3 from the right (\( x \to 3^+ \))**: \[ \lim_{x \to 3^+} h(x) = \lim_{x \to 3} \left( \frac{x^2 - 6x + 9}{x - 3} \right). \] The expression can be simplified. Notice that \( x^2 - 6x + 9 = (x - 3)^2 \): \[ \lim_{x \to 3^+} h(x) = \lim_{x \to 3} \frac{(x - 3)^2}{x - 3} = \lim_{x \to 3} (x - 3) = 0. \] Now we gather our results: - \( h(3) = 0 \) - \( \lim_{x \to 3^-} h(x) = 0 \) - \( \lim_{x \to 3^+} h(x) = 0 \) Since \( \lim_{x \to 3} h(x) = 0 \) (both left and right limits are equal) and \( h(3) \) is defined and equal to the limit, we conclude that \( h(x) \) is continuous at \( a = 3 \). The result is that the function \( h(x) \) is continuous at \( a = 3 \). --- Continuous functions can be found all around us, from the smooth curves of roller coasters to the steady flow of water in a river. The study of continuity is crucial in calculus and analysis for ensuring that functions behave predictably, making the world of mathematics both reliable and fascinating! Did you know that the concept of continuity has roots in the ancient Greeks? They debated the nature of space and time, laying the groundwork for modern mathematical concepts. Philosophers like Zeno famously pondered “Achilles and the Tortoise,” which ties into the idea of limits—an essential aspect of continuity!

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