### Continuity of a Function at a Point**Problem:** Is the function \( f(x) = \begin{cases} 3 + \sqrt{2 + x}, & x \leq 7 \\ 8 - \sqrt{x - 3}, & x > 7 \end{cases} \) continuous at \( x = 7 \)? Justify your reasoning based on the three-part definition of continuity.#### Three-Part Definition of Continuity at a Point:1. **The function is defined at the point.** For \( f(x) \) to be continuous at \( x = 7 \), \( f(7) \) must be defined.2. **The limit of the function exists as \( x \) approaches the point.** We must find \( \lim_{{x \to 7}} f(x) \) from both the left and the right.3. **The limit of the function as \( x \) approaches the point is equal to the function's value at that point.** Ensure that \( \lim_{{x \to 7}} f(x) = f(7) \).#### Applying the Definition to the Given Function:1. **Defined at \( x = 7 \):** For \( x \leq 7 \), \( f(x) = 3 + \sqrt{2 + x} \). Thus, \( f(7) = 3 + \sqrt{2 + 7} = 3 + \sqrt{9} = 3 + 3 = 6 \).2. **Limit as \( x \) approaches 7:** - **From the left** (\( x \to 7^{-} \)): \[ \lim_{{x \to 7^{-}}} f(x) = \lim_{{x \to 7^{-}}} (3 + \sqrt{2 + x}) = 3 + \sqrt{2 + 7} = 6 \] - **From the right** (\( x \to 7^{+} \)): \[ \lim_{{x \to 7^{+}}} f(x) = \lim_{{x \to 7^{+}}} (8 - \sqrt{x - 3}) = 8 - \sqrt{7

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Answered: 3+2+x, xs7 8-Jx-3, x>7 three part…

3+2+x, xs7 8-Jx-3, x>7 three part definition of continuity. Is the function f(x)= continuous at x=7? Justify your reasoning based on the

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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