let (f) be the function given above. on which of the following intervals is (f) continuous?\na ((-5,0)\nb…

Explanation:

Step1: Analyze continuity at break - points

The break - points of the piece - wise function are (x=-1), (x = 0), (x = 4). We need to check the limits at these points and also consider the domain of each sub - function. The function (y = e^{x}) is continuous for (x\leq - 1), (y=x^{3}+2x) is continuous for (-1\lt x\leq0), (y = \frac{3x}{x - 2}) is continuous for (0\lt x\leq4) except at (x = 2) (vertical asymptote), and (y=\cos(3x)) is continuous for (x>4).

Step2: Check option A

For the interval ((-5,0)), at (x=-1), (\lim_{x\rightarrow - 1^{-}}e^{x}=e^{-1}), (\lim_{x\rightarrow - 1^{+}}x^{3}+2x=(-1)^{3}+2\times(-1)=-1 - 2=-3). Since (\lim_{x\rightarrow - 1^{-}}f(x)\neq\lim_{x\rightarrow - 1^{+}}f(x)), the function is not continuous on ((-5,0)).

Step3: Check option B

For the interval ((-1,2)), at (x = 0), (\lim_{x\rightarrow0^{-}}x^{3}+2x=0), (\lim_{x\rightarrow0^{+}}\frac{3x}{x - 2}=0). But at (x = 2), the function (y=\frac{3x}{x - 2}) has a vertical asymptote. So the function is not continuous on ((-1,2)).

Step4: Check option C

For the interval ((0,3)), the function (y=\frac{3x}{x - 2}) is continuous for all (x\in(0,3)) as (x\neq2) in this interval. The sub - function (\frac{3x}{x - 2}) is a rational function and is continuous on its domain, and there are no break - points of the piece - wise function in the interval ((0,3)).

Step5: Check option D

For the interval ((2,5)), at (x = 4), (\lim_{x\rightarrow4^{-}}\frac{3x}{x - 2}=\frac{3\times4}{4 - 2}=6), (\lim_{x\rightarrow4^{+}}\cos(3x)=\cos(12)). Since (\lim_{x\rightarrow4^{-}}f(x)\neq\lim_{x\rightarrow4^{+}}f(x)), the function is not continuous on ((2,5)).

Answer:

C. ((0,3))