Question

multiple choice 2 points
question topic(s)/section(s): 1.11 defining continuity at a point
click the timer to pace yourself. some questions do not require the full 2 minutes, and others will require a little more than 2 minutes.
h(x)=\

$$\begin{cases}ae^{2x}+3&\\text{for }x < 0\\5 - cosx&\\text{for }x\\geq0\\end{cases}$$

let h be the function defined above, where a is a constant. if h is continuous at x = 0, what is the value of a?
0
2
5
1
clear my selection

Explanation:

Step1: Recall continuity condition

For a function to be continuous at \(x = 0\), \(\lim_{x
ightarrow0^{-}}h(x)=\lim_{x
ightarrow0^{+}}h(x)=h(0)\).

Step2: Calculate left - hand limit

\(\lim_{x
ightarrow0^{-}}h(x)=\lim_{x
ightarrow0^{-}}(ae^{2x}+3)=a\cdot e^{0}+3=a + 3\).

Step3: Calculate right - hand limit and function value at \(x = 0\)

\(\lim_{x
ightarrow0^{+}}h(x)=h(0)=5-\cos(0)=5 - 1=4\).

Step4: Equate left - hand and right - hand limits

Set \(a + 3=4\), then \(a=4 - 3=1\).

Answer:

D. 1