1.06a, 1.06b - Exponential functions and the number e | OCR A-Level Mathematics A

Read exponential graphs as repeated multiplication, distinguish growth from decay, and explain the special role of the number $\mathrm e$ . Use its gradient property to connect a graph's height with its rate of change.

Positive bases and repeated multiplication

Can a quantity keep growing faster while its percentage rate stays fixed? Start with 100 organisms: multiplying by $1.1$ each year gives 110, then 121. The increases are different, but the multiplier is unchanged.

You already know

The index laws give $a^{x+1}=a\,a^x$ , $a^0=1$ and $a^{-x}=1/a^x$ for $a>0$ . A geometric sequence uses a fixed multiplier between successive terms.

Let the input vary continuously instead of taking only integer values.

Positive-base exponential

An exponential function $f(x)=a^x$ has a fixed positive base $a$ and a variable exponent $x$ . For example, $2^x$ is exponential; $x^2$ is a power function.

Negative bases fail for some real inputs: $(-2)^{1/2}$ has no real value. Zero also fails to give an all-real-input exponential. Positive bases give the following complete picture.

Exponential graph features

For $a>0$ , $a^{x+1}=a\,a^x$ and $a^0=1$ . If $a\ne1$ ,

\text{domain }\mathbb R,\qquad \text{range }(0,\infty),\qquad \text{asymptote }y=0.

The graph increases for $a>1$ and decreases for $0<a<1$ . For $a=1$ , it is the constant $y=1$ , with range $\{1\}$ .

Not in the formula sheet: learn it.

The base is also the multiplier for a unit step along the horizontal axis.

Worked example 1: a growth graph

Question. Find the values of $2^x$ for $x=-2,-1,0,1,2$ , and state its main graph features.

Plan. Use negative and zero indices, then read growth from the base.

\begin{aligned} 2^{-2}&=\tfrac14,\quad 2^{-1}=\tfrac12 &&\text{reciprocals}\\ 2^0&=1,\quad 2^1=2,\quad 2^2=4 &&\text{integer powers} \end{aligned}

Why this step? Points on both sides of zero show that positive outputs can be less than 1.

Answer. The graph increases, crosses the vertical axis at $(0,1)$ , has domain $\mathbb R$ , range $(0,\infty)$ , and horizontal asymptote $y=0$ as $x\to-\infty$ .

Check. Each successive output is twice the previous output.

Change just the base to its reciprocal; the graph reverses horizontally.

Worked example 2: a decay graph

Question. Give the same values and features for $y=(1/2)^x$ .

Plan. Rewrite the function as $2^{-x}$ .

\begin{aligned} (1/2)^x&=2^{-x} &&\text{negative-index law}\\ y&=4,2,1,\tfrac12,\tfrac14 &&\text{at }x=-2,-1,0,1,2 \end{aligned}

Why this step? Replacing $x$ by $-x$ reflects the growth graph in the vertical axis.

Answer. It decreases through $(0,1)$ , with domain $\mathbb R$ , range $(0,\infty)$ and asymptote $y=0$ as $x\to\infty$ .

Check. Each unit step multiplies the output by $1/2$ ; it never makes a positive output zero.

The picture includes the separate constant case, which has no growth or decay.

Growth, decay and the constant base-one case

At integer years, $P(n)=100(1.1)^n$ is a geometric model. Using the same expression between whole years assumes smooth exponential change between observations.

Watch out

A decreasing exponential does not become negative. For $0<a<1$ , $a^x>0$ for every real $x$ ; its rate of change is negative, not its value.

Try a new reciprocal base.

The number e and its graph

One base makes the graph's gradient equal its height at every point: $\mathrm e\approx2.718281828$ .

Natural exponential

The natural exponential is $y=\mathrm e^x$ . It increases through $(0,1)$ , has domain $\mathbb R$ , range $(0,\infty)$ , and horizontal asymptote $y=0$ towards the left.

Here, gradient means the slope of the tangent: the instantaneous rate of change. We use its known value now; systematic differentiation comes later.

Height and gradient

For every real $x$ , the gradient of $y=\mathrm e^x$ is $\mathrm e^x$ .

Not in the formula sheet: learn it.

Near zero, the chord gradients $(\mathrm e^{0.01}-1)/0.01\approx1.005$ and $(\mathrm e^{-0.01}-1)/(-0.01)\approx0.995$ approach 1 from either side. These illustrate the property; two numerical checks do not prove it.

Worked example 3: the tangent at zero

Question. Find the tangent to $y=\mathrm e^x$ at $x=0$ .

Plan. Use the function value for both the height and the tangent gradient.

\begin{aligned} y&=\mathrm e^0=1 &&\text{point height}\\ m&=\mathrm e^0=1 &&\text{known gradient}\\ y-1&=1(x-0) &&\text{point-gradient form} \end{aligned}

Why this step? The same output supplies both numbers needed for a tangent line.

Answer. $y=x+1$ .

Check. The line passes through $(0,1)$ with gradient 1; it is not the whole curved graph.

Move only the point of contact to $x=1$ .

Worked example 4: a tangent through the origin

Question. Find the tangent to $y=\mathrm e^x$ at $x=1$ .

Plan. Repeat the height-and-gradient argument at the new input.

\begin{aligned} y&=\mathrm e,\quad m=\mathrm e &&\text{height equals gradient}\\ y-\mathrm e&=\mathrm e(x-1) &&\text{point-gradient form}\\ y&=\mathrm e x &&\text{simplify} \end{aligned}

Why this step? Keeping $\mathrm e$ exact makes the cancellation visible.

Answer. $y=\mathrm e x$ ; surprisingly, this tangent passes through the origin even though the exponential never does.

Check. At $x=1$ , the line gives $y=\mathrm e$ ; at $x=0$ , it gives zero.

The tangent and the curve agree at their point of contact, rather than everywhere.

The natural exponential and its tangent at the vertical-axis intercept

Use the same property at a negative input.

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1.06a, 1.06b - Exponential functions and the number e

Positive bases and repeated multiplication

The number e and its graph

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