Question

let (f) be the function given above. on which of the following intervals is (f) continuous?
a ((-5,0)
b ((-1,2)
c ((0,3)
d ((2,5)
f(x)=\begin{cases}e^{x}&\text{for }xleq - 1\\x^{3}+2x&\text{for }-1 < xleq0\\frac{3x}{x - 2}&\text{for }0 < xleq4\cos(3x)&\text{for }x>4end{cases}

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

Step3: Check option B

For the interval \((-1,2)\), at \(x = 0\), \(\lim_{x
ightarrow0^{-}}x^{3}+2x=0\), \(\lim_{x
ightarrow0^{+}}\frac{3x}{x - 2}=0\). 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 \(x
eq2\) 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 \(x = 4\), \(\lim_{x
ightarrow4^{-}}\frac{3x}{x - 2}=\frac{3\times4}{4 - 2}=6\), \(\lim_{x
ightarrow4^{+}}\cos(3x)=\cos(12)\). Since \(\lim_{x
ightarrow4^{-}}f(x)
eq\lim_{x
ightarrow4^{+}}f(x)\), the function is not continuous on \((2,5)\).

Answer:

C. \((0,3)\)