Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Sure, let's evaluate the given piecewise function step-by-step.
Given the function:
[tex]\[ f(x) = \begin{cases} -5x & \text{if } x < 1 \\ 1 & \text{if } x = 1 \\ 5x & \text{if } x > 1 \end{cases} \][/tex]
(a) [tex]\(\lim_{x \to 1^-} f(x)\)[/tex]
- This represents the limit of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 1 from the left (i.e., from values less than 1).
- When [tex]\(x < 1\)[/tex], [tex]\(f(x) = -5x\)[/tex].
Thus,
[tex]\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} -5x = -5 \cdot 1 = -5 \][/tex]
(b) [tex]\(\lim_{x \to 1^+} f(x)\)[/tex]
- This represents the limit of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 1 from the right (i.e., from values greater than 1).
- When [tex]\(x > 1\)[/tex], [tex]\(f(x) = 5x\)[/tex].
Thus,
[tex]\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5x = 5 \cdot 1 = 5 \][/tex]
(c) [tex]\(\lim_{x \to 1} f(x)\)[/tex]
- For the limit [tex]\(\lim_{x \to 1} f(x)\)[/tex] to exist, both the left-hand limit [tex]\(\lim_{x \to 1^-} f(x)\)[/tex] and the right-hand limit [tex]\(\lim_{x \to 1^+} f(x)\)[/tex] must exist and be equal.
- From (a), we have [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex].
- From (b), we have [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex].
Since these two limits are not equal, the overall limit does not exist (DNE).
(d) [tex]\(f(1)\)[/tex]
- The function value at [tex]\(x = 1\)[/tex] is given by the piecewise definition. It specifically states that [tex]\(f(x) = 1\)[/tex] when [tex]\(x = 1\)[/tex].
Thus,
[tex]\[ f(1) = 1 \][/tex]
(e) Comparison and conclusion
- Since the left-hand limit [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex] and the right-hand limit [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex] are not equal, there is a discontinuity at [tex]\(x = 1\)[/tex].
- The type of discontinuity where the left-hand and right-hand limits exist but are not equal is called a "jump discontinuity".
So, the function has a jump discontinuity at [tex]\(x = 1\)[/tex].
### Summary of Results:
(a) [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex]
(b) [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex]
(c) [tex]\(\lim_{x \to 1} f(x) = \text{DNE (Does Not Exist)}\)[/tex]
(d) [tex]\(f(1) = 1\)[/tex]
(e) The function has a jump discontinuity at [tex]\(x = 1\)[/tex].
Given the function:
[tex]\[ f(x) = \begin{cases} -5x & \text{if } x < 1 \\ 1 & \text{if } x = 1 \\ 5x & \text{if } x > 1 \end{cases} \][/tex]
(a) [tex]\(\lim_{x \to 1^-} f(x)\)[/tex]
- This represents the limit of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 1 from the left (i.e., from values less than 1).
- When [tex]\(x < 1\)[/tex], [tex]\(f(x) = -5x\)[/tex].
Thus,
[tex]\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} -5x = -5 \cdot 1 = -5 \][/tex]
(b) [tex]\(\lim_{x \to 1^+} f(x)\)[/tex]
- This represents the limit of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 1 from the right (i.e., from values greater than 1).
- When [tex]\(x > 1\)[/tex], [tex]\(f(x) = 5x\)[/tex].
Thus,
[tex]\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5x = 5 \cdot 1 = 5 \][/tex]
(c) [tex]\(\lim_{x \to 1} f(x)\)[/tex]
- For the limit [tex]\(\lim_{x \to 1} f(x)\)[/tex] to exist, both the left-hand limit [tex]\(\lim_{x \to 1^-} f(x)\)[/tex] and the right-hand limit [tex]\(\lim_{x \to 1^+} f(x)\)[/tex] must exist and be equal.
- From (a), we have [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex].
- From (b), we have [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex].
Since these two limits are not equal, the overall limit does not exist (DNE).
(d) [tex]\(f(1)\)[/tex]
- The function value at [tex]\(x = 1\)[/tex] is given by the piecewise definition. It specifically states that [tex]\(f(x) = 1\)[/tex] when [tex]\(x = 1\)[/tex].
Thus,
[tex]\[ f(1) = 1 \][/tex]
(e) Comparison and conclusion
- Since the left-hand limit [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex] and the right-hand limit [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex] are not equal, there is a discontinuity at [tex]\(x = 1\)[/tex].
- The type of discontinuity where the left-hand and right-hand limits exist but are not equal is called a "jump discontinuity".
So, the function has a jump discontinuity at [tex]\(x = 1\)[/tex].
### Summary of Results:
(a) [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex]
(b) [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex]
(c) [tex]\(\lim_{x \to 1} f(x) = \text{DNE (Does Not Exist)}\)[/tex]
(d) [tex]\(f(1) = 1\)[/tex]
(e) The function has a jump discontinuity at [tex]\(x = 1\)[/tex].
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
