Exponential functions are a family of functions that take the form of \(f(x) = a^x\), where a is some constant above 0).
Here is what \(f(x) = 2^x\) looks like:
And what \(f(x) = 3^x\) looks like:
Note that all expontial functions of the exact form \(f(x) = a^x\) will always have coordinate of (0, 1) and (1, \(a\)), and feature an asymptote at \(y = 0\)
Functions of \(f(x) = a^x+b\) represent that graph \(f(x) = a^x\) that have been translated \(b\) units in the positive y-direction. If \(b\) is negative, the graph is translated downwards:
\(f(x) = 2^x\)
\(f(x) = 2^x + 2\)
\(f(x) = 2^x - 1\)
\(f(x) = 2^x - 2\)
Functions of \(f(x) = a^{x+c}+b\) represent that graph \(f(x) = a^x + b\) that have been translated \(c\) units in the negative x-direction. If \(b\) is negative, the graph is translated downwards:
\(f(x) = 1.7^x\)
\(f(x) = 1.7^{x+1}\)
\(f(x) = 1.7^{x-2}\)
\(f(x) = 1.7^{x-2}+2\)
If you havn't realised, this function class follows most with translations and dilations. The function \(f(x) = a^{bc+x} + d\) represent the graph \(f(x) = a^{x+c} + d\) in which the value for x is multiplied by \(b\).
\(f(x) = 3^{x-2} + 2\)
\(f(x) = 3^{-x-2} + 2\)
\(f(x) = 3^{2x-2} + 2\)
\(f(x) = 3^{0.5x - 2} + 2\)
The function \(f(x) = g(a^{bc+x} + d\)) represent the graph \(f(x) = a^{bx+c} + d\) in which the value for \(f(x)\) is multiplied by \(g\).
\(f(x) = [ 2^{1.5x-2}]\)
\(f(x) = -[ 2^{1.5x-2}]\)
\(f(x) = 2[ 2^{1.5x-2}]\)
\(f(x) = 0.5[ 2^{1.5x-2}]\)
Now baby, let me introduce you to eulers number, \(e\). This number is a very significant constant that begins with 2.71828, and never ends. This number pops up in many different contexts within mathmatics, and is features in topics such as calculus, finance, number series of factorials.
This is what the \(f(x) = e^x\) looks like:
Pretty boring... right? WRONG. This equation is very interesting because: it is it's own derivitive:
\(\displaystyle \frac{d}{dx} e^x = e^x\)
And so of course:
\(\displaystyle \int e^x \ dx = e^x + c \)
We can show this happens using this limit, whilst incrementing \(a\) to approach \(e\)
\(\displaystyle \boxed{\lim_{h \rightarrow 0} \frac{a^h -1}{h}} \)
\(\displaystyle \begin{aligned} \lim_{h \rightarrow 0} \frac{1.05^h -1}{h} &= 0.0488 \\[12pt] \lim_{h \rightarrow 0} \frac{1.2^h -1}{h} &= 0.1823 \\[12pt] \lim_{h \rightarrow 0} \frac{1.6^h -1}{h} &= 0.4700 \\[12pt] \lim_{h \rightarrow 0} \frac{2^h -1}{h} &= 0.6931 \\[12pt] \lim_{h \rightarrow 0} \frac{2.4^h -1}{h} &= 0.8755 \\[12pt] \lim_{h \rightarrow 0} \frac{2.8^h -1}{h} &= 0.1.0296 \\[12pt] \lim_{h \rightarrow 0} \frac{2.7^h -1}{h} &= 0.9933 \\[12pt] \lim_{h \rightarrow 0} \frac{2.714^h -1}{h} &= 0.9984 \\[12pt] \lim_{h \rightarrow 0} \frac{e^h -1}{h} &= 1 \end{aligned} \)
This indicates that \(e\) is a unique number for which it's exponent's derivative is equal to it's exponent.
The chain rule offers guidance for the derivative and integration of an exponent of \(e\) involving some \(f(x)\).
Where \(y = e^{f(x)}\):
\(\displaystyle \frac{dy}{dx} = f'(x) e^{f(x)}\)
\(\int f'(x) e^{f(x)} \ dx = e^{f(x)} + c\)
1) Calculate \(\displaystyle \frac{d}{dx} (e^{x^3-2x} + 2)\)
\( \ \ \ \ \ \ = (3x^2 -2)e^{x^3-2x}\)
2) Calculate \(\displaystyle \frac{d}{dx} (3x^e - 2x -2e^{-2x}) \)
\( \ \ \ \ \ \ = 3ex^{e-1} -2 +4e^{-2x} \)
3) Calculate \(\displaystyle \int e^{-7x} \ dx\)
\( \ \ \ \ \ \ \displaystyle \begin{aligned} &= - \frac{1}{7} \int -7e^{-7x} \ dx \\[5pt] &= - \frac{1}{7} e^{-7x} + c \end{aligned} \)
4) Calculate \(\displaystyle \int 9x^2 e^{x^3} + 2e \ dx\)
\( \ \ \ \ \ \ \displaystyle \begin{aligned} &= 3 \int 3x^2 e^{x^3} \ dx + \int 2e \ dx \\[5pt] &= 3e^{x^3} + 2ex + c \end{aligned} \)
The format of exponential functions involving \(e\) can be used to model sitations involving exponential growth and decay. We can, like all derivitive functions, use integration to determine the total/net change.
An exponential functions rate of change is directly proportional to the value of the function. The formula for population, \(P\), following exponential growth and decay is:
\(P = P_0 e^{kt}\)
\(\frac{dP}{dt} = kP\)
Where \(P_0\) is the initial population, \(k\) is the growth rate, such that:
5) A population of rabits is killed off by some foxes such that the population is decreasing at 11% per year. If the population originally consists of 1200 rabbits, how many rabbits will be alive after 28 months?
\(\begin{aligned} \frac{dP}{dt} &= -0.11P \\[5pt] P &= 1200e^{-0.11t} \\[5pt] P_{t = 28} &= 1200e^{-0.11 \times \frac{28}{12}} \\[5pt] &= 928.35 \\[5pt] &\approx 928 \end{aligned} \)
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