Exponential functionsAQA A-Level Maths: Revision notes

An exponential function has the variable in the power: $f(x)=a^x$ with $a>0$ (and $a\ne1$). Key features of its graph:

• It passes through $(0,1)$, because $a^0=1$.
• It is always positive: $a^x>0$ for every $x$, so the graph never meets the $x$-axis. The $x$-axis ($y=0$) is a horizontal asymptote.
• If $a>1$ the graph increases (growth), getting very steep for large $x$ and approaching $0$ as $x\to-\infty$. If $0<a<1$ it decreases (decay).
• $\left(\frac1a\right)^x=a^{-x}$, so $y=a^{-x}$ is the reflection of $y=a^x$ in the $y$-axis. Example: $2^x=16$ gives $x=4$ and $2^x=\frac18$ gives $x=-3$.

The number $e\approx2.718$ is the special base for which the gradient of $y=a^x$ at $(0,1)$ is exactly $1$. The function $y=e^x$ (the exponential function) has the same properties as other $a^x$ with $a>1$: through $(0,1)$, always positive, asymptote $y=0$, increasing. Its key property is that the gradient at any point equals the $y$-value: $\frac{d}{dx}(e^x)=e^x$. At $x=0$ the gradient is $1$; at $x=1$ it is $e$.

For any constant $k$, $\frac{d}{dx}\left(e^{kx}\right)=ke^{kx}.$ More generally, $\frac{d}{dx}\left(Ae^{kx}\right)=Ake^{kx}$. The constant $k$ multiplies the gradient. Examples: $\frac{d}{dx}(e^{2x})=2e^{2x}$; $\frac{d}{dx}(5e^{-2x})=-10e^{-2x}$; $\frac{d}{dx}(20+70e^{-0.05t})=-3.5e^{-0.05t}$ (the constant $20$ differentiates to $0$). Worked example: the tangent to $y=e^{2x}$ at $x=\frac12$. The point is $\left(\frac12,e\right)$ and the gradient is $2e$, so $y-e=2e\left(x-\frac12\right)$, which simplifies to $y=2ex$.

Since $\frac{d}{dx}(e^{kx})=ke^{kx}$, the gradient of $y=Ae^{kx}$ is $k$ times $y$. So the rate of change is proportional to the quantity itself: $\frac{dy}{dx}=ky$.

• $k>0$: growth (population, compound interest), ever faster.
• $k<0$: decay (radioactivity, cooling towards a limit), ever slower. Example: $N=500e^{0.4t}$ has $\frac{dN}{dt}=200e^{0.4t}=0.4N$. At $t=5$ the rate is $200e^2\approx1480$ bacteria per hour. For Newton cooling, $T=20+70e^{-0.05t}$ gives $\frac{dT}{dt}=-0.05(T-20)$: the cooling rate is proportional to the excess over room temperature. The model predicts $T\to20$ as $t\to\infty$; real models have limits (food, space) that an exponential ignores.