Which function has a removable discontinuity?

Reasoning:

The denominator in choice A factors to (x-2)(x+1). We have a pair of "x-2" terms that cancel leaving us with 1/(x+1). We cannot plug x = -1 into the simplified result because it leads the denominator x+1 to turn into 0.

Also, we cannot plug x = 2 into p(x) for choice A either. It doesn't lead to issues for 1/(x+1) but the original function before we simplify will have x = 2 lead to a division by zero error.

A vertical asymptote is located at x = -1

The removable discontinuity happens when x = 2, so there will be a hole here.

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answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: