QUESTION IMAGE
Question
for each situation, answer the questions.
- the following graph displays the exponential function ( f(x)=2^{x}+3 ) with the appropriate asymptote.
what is the right - end behavior and the left - end behavior for this function written with infinity notation?
a) right - end behavior:
b) left - end behavior:
end behavior of functions
A) Right - end behavior
To determine the right - end behavior of the function \(f(x)=2e^{x}+3\), we analyze the limit as \(x\to+\infty\).
The exponential function \(y = e^{x}\) has the property that as \(x\to+\infty\), \(e^{x}\to+\infty\).
For the function \(f(x)=2e^{x}+3\), when \(x\) approaches \(+\infty\) (right - end), we consider the limit \(\lim_{x\to+\infty}(2e^{x}+3)\).
Since \(\lim_{x\to+\infty}e^{x}=+\infty\), then \(\lim_{x\to+\infty}2e^{x}=2\times\lim_{x\to+\infty}e^{x}=+\infty\). And \(\lim_{x\to+\infty}3 = 3\). So \(\lim_{x\to+\infty}(2e^{x}+3)=+\infty\). In terms of end - behavior notation, as \(x\to+\infty\), \(f(x)\to+\infty\).
B) Left - end behavior
To determine the left - end behavior of the function \(f(x)=2e^{x}+3\), we analyze the limit as \(x\to-\infty\).
The exponential function \(y = e^{x}\) has the property that as \(x\to-\infty\), \(e^{x}\to0\) (because \(e^{x}=\frac{1}{e^{\vert x\vert}}\) and as \(\vert x\vert\to+\infty\) when \(x\to-\infty\), \(e^{\vert x\vert}\to+\infty\), so \(\frac{1}{e^{\vert x\vert}}\to0\)).
For the function \(f(x)=2e^{x}+3\), when \(x\) approaches \(-\infty\) (left - end), we consider the limit \(\lim_{x\to-\infty}(2e^{x}+3)\).
We know that \(\lim_{x\to-\infty}e^{x}=0\), so \(\lim_{x\to-\infty}2e^{x}=2\times\lim_{x\to-\infty}e^{x}=0\). And \(\lim_{x\to-\infty}3 = 3\). So \(\lim_{x\to-\infty}(2e^{x}+3)=0 + 3=3\). In terms of end - behavior notation, as \(x\to-\infty\), \(f(x)\to3\).
A) Answer: As \(x\to+\infty\), \(f(x)\to+\infty\)
B) Answer: As \(x\to-\infty\), \(f(x)\to3\)
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To determine the left - end behavior of the function \(f(x)=2e^{x}+3\), we analyze the limit as \(x\to-\infty\).
The exponential function \(y = e^{x}\) has the property that as \(x\to-\infty\), \(e^{x}\to0\) (because \(e^{x}=\frac{1}{e^{\vert x\vert}}\) and as \(\vert x\vert\to+\infty\) when \(x\to-\infty\), \(e^{\vert x\vert}\to+\infty\), so \(\frac{1}{e^{\vert x\vert}}\to0\)).
For the function \(f(x)=2e^{x}+3\), when \(x\) approaches \(-\infty\) (left - end), we consider the limit \(\lim_{x\to-\infty}(2e^{x}+3)\).
We know that \(\lim_{x\to-\infty}e^{x}=0\), so \(\lim_{x\to-\infty}2e^{x}=2\times\lim_{x\to-\infty}e^{x}=0\). And \(\lim_{x\to-\infty}3 = 3\). So \(\lim_{x\to-\infty}(2e^{x}+3)=0 + 3=3\). In terms of end - behavior notation, as \(x\to-\infty\), \(f(x)\to3\).