Answer :
To determine on which interval the function [tex]\( f(x) \)[/tex] is continuous, we need to analyze each given interval and check for continuity throughout.
Option (A) [tex]\((-5,0)\)[/tex]:
Let's split this interval into two parts and analyze it:
- On [tex]\((-5, -1)\)[/tex], [tex]\( f(x) = e^x \)[/tex]. The function [tex]\( e^x \)[/tex] is continuous for all [tex]\( x \)[/tex].
- On [tex]\((-1, 0)\)[/tex], [tex]\( f(x) = x^3 + 2x \)[/tex]. The function [tex]\( x^3 + 2x \)[/tex] is also continuous for all [tex]\( x \)[/tex].
We need to check the continuity at the boundary [tex]\( x = -1 \)[/tex]:
- As [tex]\( x \)[/tex] approaches [tex]\(-1\)[/tex] from the left, [tex]\( \lim_{{x \to -1^-}} f(x) = e^{-1} \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\(-1\)[/tex] from the right, [tex]\( \lim_{{x \to -1^+}} f(x) = (-1)^3 + 2(-1) = -1 - 2 = -3 \)[/tex].
Since [tex]\( e^{-1} \neq -3 \)[/tex], [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = -1 \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is not continuous on the interval [tex]\((-5,0)\)[/tex].
Option (B) [tex]\((-1,2)\)[/tex]:
We need to analyze this interval in parts:
- On [tex]\((-1, 0)\)[/tex], [tex]\( f(x) = x^3 + 2x \)[/tex]. This function is continuous.
- On [tex]\((0, 2)\)[/tex], [tex]\( f(x) = \frac{5x}{x-2} \)[/tex]. This function has a potential issue at [tex]\( x = 2 \)[/tex], but [tex]\( x = 2 \)[/tex] is not included in the interval [tex]\((-1, 2)\)[/tex].
Checks at the boundaries:
- As [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex] from the left, [tex]\( \lim_{{x \to 0^-}} f(x) = 2(0) = 0 \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex] from the right, [tex]\( \lim_{{x \to 0^+}} f(x) = \frac{5 \cdot 0}{0-2} = 0 \)[/tex].
Both limits are equal, so [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 0 \)[/tex]. No other problematic points exist in [tex]\((-1, 2)\)[/tex].
Thus, [tex]\( f(x) \)[/tex] is continuous on the interval [tex]\((-1,2)\)[/tex].
Option (C) [tex]\((0,3)\)[/tex]:
Within this interval:
- On [tex]\((0, 3)\)[/tex], [tex]\( f(x) = \frac{5x}{x-2} \)[/tex] but we need to check the point [tex]\( x = 2 \)[/tex]:
- As [tex]\( x \)[/tex] approaches [tex]\( 2 \)[/tex] from either side, [tex]\( f(x) = \frac{5(2)}{2-2} = \frac{10}{0} \)[/tex], which is undefined.
Thus, [tex]\( f(x) \)[/tex] is not continuous on the interval [tex]\((0,3)\)[/tex] because of the discontinuity at [tex]\( x = 2 \)[/tex].
Option (D) [tex]\((2,5)\)[/tex]:
This interval splits into:
- On [tex]\((2, 4)\)[/tex], [tex]\( f(x) = \frac{5x}{x-2} \)[/tex]. This function is undefined at [tex]\( x = 2 \)[/tex].
- On [tex]\((4, 5)\)[/tex], [tex]\( f(x) = \cos(3x) \)[/tex], which is continuous.
We check continuity at the boundary [tex]\( x = 4 \)[/tex]:
- As [tex]\( x \)[/tex] approaches [tex]\( 4 \)[/tex] from the left, [tex]\( \lim_{{x \to 4^-}} f(x) = \frac{5(4)}{4-2} = 10 \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( 4 \)[/tex] from the right, [tex]\( \lim_{{x \to 4^+}} f(x) = \cos(12) \)[/tex].
Since [tex]\( 10 \neq \cos(12) \)[/tex], [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = 4 \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is not continuous on the interval [tex]\((2,5)\)[/tex].
Given all these observations, the function [tex]\( f \)[/tex] is continuous in the interval:
- [tex]\(\boxed{(-1,2)}\)[/tex]
Option (A) [tex]\((-5,0)\)[/tex]:
Let's split this interval into two parts and analyze it:
- On [tex]\((-5, -1)\)[/tex], [tex]\( f(x) = e^x \)[/tex]. The function [tex]\( e^x \)[/tex] is continuous for all [tex]\( x \)[/tex].
- On [tex]\((-1, 0)\)[/tex], [tex]\( f(x) = x^3 + 2x \)[/tex]. The function [tex]\( x^3 + 2x \)[/tex] is also continuous for all [tex]\( x \)[/tex].
We need to check the continuity at the boundary [tex]\( x = -1 \)[/tex]:
- As [tex]\( x \)[/tex] approaches [tex]\(-1\)[/tex] from the left, [tex]\( \lim_{{x \to -1^-}} f(x) = e^{-1} \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\(-1\)[/tex] from the right, [tex]\( \lim_{{x \to -1^+}} f(x) = (-1)^3 + 2(-1) = -1 - 2 = -3 \)[/tex].
Since [tex]\( e^{-1} \neq -3 \)[/tex], [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = -1 \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is not continuous on the interval [tex]\((-5,0)\)[/tex].
Option (B) [tex]\((-1,2)\)[/tex]:
We need to analyze this interval in parts:
- On [tex]\((-1, 0)\)[/tex], [tex]\( f(x) = x^3 + 2x \)[/tex]. This function is continuous.
- On [tex]\((0, 2)\)[/tex], [tex]\( f(x) = \frac{5x}{x-2} \)[/tex]. This function has a potential issue at [tex]\( x = 2 \)[/tex], but [tex]\( x = 2 \)[/tex] is not included in the interval [tex]\((-1, 2)\)[/tex].
Checks at the boundaries:
- As [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex] from the left, [tex]\( \lim_{{x \to 0^-}} f(x) = 2(0) = 0 \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex] from the right, [tex]\( \lim_{{x \to 0^+}} f(x) = \frac{5 \cdot 0}{0-2} = 0 \)[/tex].
Both limits are equal, so [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 0 \)[/tex]. No other problematic points exist in [tex]\((-1, 2)\)[/tex].
Thus, [tex]\( f(x) \)[/tex] is continuous on the interval [tex]\((-1,2)\)[/tex].
Option (C) [tex]\((0,3)\)[/tex]:
Within this interval:
- On [tex]\((0, 3)\)[/tex], [tex]\( f(x) = \frac{5x}{x-2} \)[/tex] but we need to check the point [tex]\( x = 2 \)[/tex]:
- As [tex]\( x \)[/tex] approaches [tex]\( 2 \)[/tex] from either side, [tex]\( f(x) = \frac{5(2)}{2-2} = \frac{10}{0} \)[/tex], which is undefined.
Thus, [tex]\( f(x) \)[/tex] is not continuous on the interval [tex]\((0,3)\)[/tex] because of the discontinuity at [tex]\( x = 2 \)[/tex].
Option (D) [tex]\((2,5)\)[/tex]:
This interval splits into:
- On [tex]\((2, 4)\)[/tex], [tex]\( f(x) = \frac{5x}{x-2} \)[/tex]. This function is undefined at [tex]\( x = 2 \)[/tex].
- On [tex]\((4, 5)\)[/tex], [tex]\( f(x) = \cos(3x) \)[/tex], which is continuous.
We check continuity at the boundary [tex]\( x = 4 \)[/tex]:
- As [tex]\( x \)[/tex] approaches [tex]\( 4 \)[/tex] from the left, [tex]\( \lim_{{x \to 4^-}} f(x) = \frac{5(4)}{4-2} = 10 \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( 4 \)[/tex] from the right, [tex]\( \lim_{{x \to 4^+}} f(x) = \cos(12) \)[/tex].
Since [tex]\( 10 \neq \cos(12) \)[/tex], [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = 4 \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is not continuous on the interval [tex]\((2,5)\)[/tex].
Given all these observations, the function [tex]\( f \)[/tex] is continuous in the interval:
- [tex]\(\boxed{(-1,2)}\)[/tex]
