3 Exponential growth

If $a>1$ then it can be shown that, no matter how large $K$ is:

$\frac{a^x}{x^K}\to \infty \text{as}x\to \infty$

That is, if $K$ is fixed (though chosen as large as desired) then eventually, as $x$ increases, $a^x$ will become larger than the value $x^K$ provided $a>1$ . The growth of $a^x$ as $x$ increases is called exponential growth .

Task!

A function $f\left(x\right)$ grows exponentially and is such that $f\left(0\right)=1$ and $f\left(2\right)=4$ . Find the exponential curve that fits through these points. Assume the function is $f\left(x\right)={\text{e}}^{kx}$ where $k$ is to be determined from the given information. Find the value of $k$ .

First, find $f\left(0\right)$ and $f\left(2\right)$ by substituting in $f\left(x\right)={\text{e}}^{kx}$ :

By trying values of $k$ : $0.6,0.7,0.8,$ find the value such that ${\text{e}}^{2k}\approx 4$ :

Answer

${\text{e}}^{2\left(0.6\right)}=3.32$ (too low) ${\text{e}}^{2\left(0.7\right)}=4.055$ (too high)

Now try values of $k$ : $k=0.67,0.68,0.69$ :

Answer

${\text{e}}^{2\left(0.67\right)}=3.819$ (low) ${\text{e}}^{2\left(0.68\right)}=3.896$ (low) ${\text{e}}^{2\left(0.69\right)}=3.975$  (low) Next try values of $k=0.691,0.692$ :

Answer

${\text{e}}^{2\left(0.691\right)}=3.983$ , (low) ${\text{e}}^{2\left(0.692\right)}=3.991$  (low) ${\text{e}}^{2\left(0.693\right)}=3.999$  (low) Finally, state the exponential function with $k$ to 3 d.p. which most closely satisfies the conditions:

Answer