Essentially, the whole point of this section is that \(e\) is a number, so we treat it like any other number. You donβt need to know what the decimal is, but it will come in handy to know that itβs a little bit less than 3.
The rate comes from the thing thatβs raised to the \(x\text{,}\) which in this case is \(e\text{.}\) That means that \(e=1+\text{rate(decimal)}\text{.}\) Solving for the rate, we get that the rate (as a decimal) is \(e-1\text{.}\) To convert to a percentage, we just need to mulitply by 100. So, the rate is \((e-1)100\)%. Since \(e\) is approximately 3, \(e-1\gt0\text{,}\) so this is a growth rate.
The rate comes from the thing thatβs raised to the \(x\text{.}\) In this case, we need to separate the \(x\) from the negative thatβs in the exponent. Since \(e^{-x}=(e^{-1})^x\text{,}\) we have that the thing thatβs raised to the \(x\) is \(e^{-1}\text{,}\) which we can write as \(\frac{1}{e}\text{.}\) That means that \(\frac{1}{e}=1+\text{rate(decimal)}\text{.}\) Solving for the rate, we get that the rate (as a decimal) is \(\frac{1}{e}-1\text{.}\) To convert to a percentage, we just need to mulitply by 100. So, the rate is \(\left(\frac{1}{e}-1\right)100\)%. Since \(e\) is approximately 3,\(\frac{1}{e}\) is approximately \(\frac{1}{3}\text{,}\) so \(\frac{1}{e}-1\lt0\text{,}\) so this is a decay rate.
The rate comes from the thing thatβs raised to the \(x\text{,}\) which in this case is \(e\text{.}\) That means that \(e=1+\text{rate(decimal)}\text{.}\) Solving for the rate, we get that the rate (as a decimal) is \(e-1\text{.}\) To convert to a percentage, we just need to mulitply by 100. So, the rate is \((e-1)100\)%. Since \(e\) is approximately 3, \(e-1\gt0\text{,}\) so this is a growth rate.
The rate comes from the thing thatβs raised to the \(x\text{.}\) In this case, we need to separate the \(x\) from the 13 thatβs in the exponent. Since \(e^{13x}=(e^{13})^x\text{,}\) we have that the thing thatβs raised to the \(x\) is \(e^{13}\text{.}\) That means that \(e^{13}=1+\text{rate(decimal)}\text{.}\) Solving for the rate, we get that the rate (as a decimal) is \(e^{13}-1\text{.}\) To convert to a percentage, we just need to mulitply by 100. So, the rate is \((e^{13}-1)100\)%. Since \(e\) is approximately 3, \(e^{13}-1\gt0\text{,}\) so this is a growth rate.
The rate comes from the thing thatβs raised to the \(x\text{.}\) In this case, we need to separate the \(x\) from the \(-2\) thatβs in the exponent. Since \(e^{-2x}=(e^{-2})^x\text{,}\) we have that the thing thatβs raised to the \(x\) is \(e^{-2}\text{,}\) which we can write as \(\frac{1}{e^2}\text{.}\) That means that \(\frac{1}{e^2}=1+\text{rate(decimal)}\text{.}\) Solving for the rate, we get that the rate (as a decimal) is \(\frac{1}{e^2}-1\text{.}\) To convert to a percentage, we just need to mulitply by 100. So, the rate is \(\left(\frac{1}{e^2}-1\right)100\)%. Since \(e\) is approximately 3,\(\frac{1}{e^2}\) is approximately \(\frac{1}{9}\text{,}\) so \(\frac{1}{e^2}-1\lt0\text{,}\) so this is a decay rate.
Letβs find the end behavior of the exponential function \(f(x)=0.7e^{-x}\text{.}\) Since \(e^{-1}\lt 1\text{,}\) this function represents exponential decay, so the graph looks like the picture below:
For \(g(x)=5e^{x+4}-8\text{,}\) we have exponential growth, like the problem above, except that this function is shifted down by 8. So, we have the following end behavior: