Answer :
To evaluate the given piecewise function [tex]\( f(x) \)[/tex] at [tex]\( x = 6 \)[/tex], we need to consider the values approaching [tex]\( x = 6 \)[/tex] from the left and the right, and compare these values as well as the function's definition at [tex]\( x = 6 \)[/tex].
Given the piecewise function:
[tex]\[ f(x) = \begin{cases} x^2 + 5x + 4 & \text{if } x < 6 \\ 24 & \text{if } x = 6 \\ -5x + 1 & \text{if } x > 6 \end{cases} \][/tex]
### (a) Evaluating [tex]\(\lim_{x \to 6^{-}} f(x)\)[/tex]
For [tex]\( x < 6 \)[/tex], [tex]\( f(x) = x^2 + 5x + 4 \)[/tex].
To find [tex]\(\lim_{x \to 6^{-}} f(x)\)[/tex], we substitute [tex]\( x = 6 \)[/tex] into the expression [tex]\( x^2 + 5x + 4 \)[/tex]:
[tex]\[ \lim_{x \to 6^{-}} f(x) = 6^2 + 5(6) + 4 = 36 + 30 + 4 = 70 \][/tex]
So,
[tex]\[ \lim_{x \to 6^{-}} f(x) = 70 \][/tex]
### (b) Evaluating [tex]\(\lim_{x \to 6^{+}} f(x)\)[/tex]
For [tex]\( x > 6 \)[/tex], [tex]\( f(x) = -5x + 1 \)[/tex].
To find [tex]\(\lim_{x \to 6^{+}} f(x)\)[/tex], we substitute [tex]\( x = 6 \)[/tex] into the expression [tex]\( -5x + 1 \)[/tex]:
[tex]\[ \lim_{x \to 6^{+}} f(x) = -5(6) + 1 = -30 + 1 = -29 \][/tex]
So,
[tex]\[ \lim_{x \to 6^{+}} f(x) = -29 \][/tex]
### (c) Evaluating the nature of the discontinuity at [tex]\( x = 6 \)[/tex]
- The left-hand limit as [tex]\( x \)[/tex] approaches 6: [tex]\( \lim_{x \to 6^{-}} f(x) = 70 \)[/tex]
- The right-hand limit as [tex]\( x \)[/tex] approaches 6: [tex]\( \lim_{x \to 6^{+}} f(x) = -29 \)[/tex]
- The value of the function at [tex]\( x = 6 \)[/tex]: [tex]\( f(6) = 24 \)[/tex]
Since [tex]\(\lim_{x \to 6^{-}} f(x) \neq \lim_{x \to 6^{+}} f(x)\)[/tex] (i.e., 70 ≠ -29), the limit does not exist (they are not equal). This indicates a jump discontinuity at [tex]\( x = 6 \)[/tex].
Therefore, the function [tex]\( f(x) \)[/tex] has a jump discontinuity at [tex]\( x = 6 \)[/tex].
Summarizing:
(a) [tex]\(\lim_{x \rightarrow 6^{-}} f(x) = 70\)[/tex]
(b) [tex]\(\lim_{x \rightarrow 6^{+}} f(x) = -29\)[/tex]
(c) The function has a jump discontinuity at [tex]\( x = 6 \)[/tex].
Given the piecewise function:
[tex]\[ f(x) = \begin{cases} x^2 + 5x + 4 & \text{if } x < 6 \\ 24 & \text{if } x = 6 \\ -5x + 1 & \text{if } x > 6 \end{cases} \][/tex]
### (a) Evaluating [tex]\(\lim_{x \to 6^{-}} f(x)\)[/tex]
For [tex]\( x < 6 \)[/tex], [tex]\( f(x) = x^2 + 5x + 4 \)[/tex].
To find [tex]\(\lim_{x \to 6^{-}} f(x)\)[/tex], we substitute [tex]\( x = 6 \)[/tex] into the expression [tex]\( x^2 + 5x + 4 \)[/tex]:
[tex]\[ \lim_{x \to 6^{-}} f(x) = 6^2 + 5(6) + 4 = 36 + 30 + 4 = 70 \][/tex]
So,
[tex]\[ \lim_{x \to 6^{-}} f(x) = 70 \][/tex]
### (b) Evaluating [tex]\(\lim_{x \to 6^{+}} f(x)\)[/tex]
For [tex]\( x > 6 \)[/tex], [tex]\( f(x) = -5x + 1 \)[/tex].
To find [tex]\(\lim_{x \to 6^{+}} f(x)\)[/tex], we substitute [tex]\( x = 6 \)[/tex] into the expression [tex]\( -5x + 1 \)[/tex]:
[tex]\[ \lim_{x \to 6^{+}} f(x) = -5(6) + 1 = -30 + 1 = -29 \][/tex]
So,
[tex]\[ \lim_{x \to 6^{+}} f(x) = -29 \][/tex]
### (c) Evaluating the nature of the discontinuity at [tex]\( x = 6 \)[/tex]
- The left-hand limit as [tex]\( x \)[/tex] approaches 6: [tex]\( \lim_{x \to 6^{-}} f(x) = 70 \)[/tex]
- The right-hand limit as [tex]\( x \)[/tex] approaches 6: [tex]\( \lim_{x \to 6^{+}} f(x) = -29 \)[/tex]
- The value of the function at [tex]\( x = 6 \)[/tex]: [tex]\( f(6) = 24 \)[/tex]
Since [tex]\(\lim_{x \to 6^{-}} f(x) \neq \lim_{x \to 6^{+}} f(x)\)[/tex] (i.e., 70 ≠ -29), the limit does not exist (they are not equal). This indicates a jump discontinuity at [tex]\( x = 6 \)[/tex].
Therefore, the function [tex]\( f(x) \)[/tex] has a jump discontinuity at [tex]\( x = 6 \)[/tex].
Summarizing:
(a) [tex]\(\lim_{x \rightarrow 6^{-}} f(x) = 70\)[/tex]
(b) [tex]\(\lim_{x \rightarrow 6^{+}} f(x) = -29\)[/tex]
(c) The function has a jump discontinuity at [tex]\( x = 6 \)[/tex].
