ME1, ME8, ME9 - Exponential functions and the number e | OCR A-Level Mathematics B (MEI)

Connect exponential graphs with repeated multiplication and interpret their intercepts, ranges and asymptotes. Use the special gradient of $\mathrm e^{kx}$ to explain when an exponential is a sensible model for change.

Positive bases and repeated multiplication

If a device loses one fifth of its remaining charge each hour, does it lose the same amount every hour? Starting from 100 units gives 80, then 64: the losses are 20 and 16, although the multiplier stays $0.8$ .

You already know

For $a>0$ , $a^{x+1}=a\,a^x$ , $a^0=1$ and $a^{-x}=1/a^x$ . A geometric sequence multiplies by the same ratio at each step.

The same multiplication pattern can describe a continuous curve.

Exponential function

An exponential function $y=a^x$ has a fixed positive base and a variable exponent. For example, $3^x$ is exponential; $x^3$ is not.

A non-positive base does not give the required real output for every real input: $(-3)^{1/2}$ is not real, and a zero base fails at negative exponents.

Positive-base graph features

For $a>0$ , $a^0=1$ . When $a\ne1$ ,

\text{domain }\mathbb R,\qquad\text{range }(0,\infty),\qquad\text{horizontal 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.

To sketch, establish the direction and intercept before joining a few exact points.

Worked example 1: building a growth sketch

Question. Find the points on $y=3^x$ at $x=-1,0,1,2$ , and state its range and asymptote.

Plan. Evaluate powers and use positivity to locate the whole curve.

\begin{aligned} 3^{-1}&=\tfrac13,\quad 3^0=1 &&\text{negative and zero indices}\\ 3^1&=3,\quad 3^2=9 &&\text{positive indices} \end{aligned}

Why this step? Points to the left of zero show that positive outputs need not exceed 1.

Answer. The points are $(-1,1/3),(0,1),(1,3),(2,9)$ . The curve increases, has range $(0,\infty)$ , and approaches $y=0$ as $x\to-\infty$ .

Check. Each output is three times the previous one.

Change only the base to its reciprocal, which reverses the input direction.

Worked example 2: building a decay sketch

Question. Give the corresponding points, range and asymptote for $y=(1/3)^x$ .

Plan. Write $(1/3)^x=3^{-x}$ and reverse the powers.

\begin{aligned} (1/3)^{-1}&=3,\quad (1/3)^0=1 &&\text{first two inputs}\\ (1/3)^1&=\tfrac13,\quad (1/3)^2=\tfrac19 &&\text{next two inputs} \end{aligned}

Why this step? The reciprocal base reflects the graph in the vertical axis, without making any output negative.

Answer. The points are $(-1,3),(0,1),(1,1/3),(2,1/9)$ . The curve decreases, has range $(0,\infty)$ , and approaches $y=0$ as $x\to\infty$ .

Check. The multiplier is $1/3$ at each step.

The two reciprocal bases give the reflected shapes below; the dashed line shows the constant exception.

Growth, decay and the separate base-one graph

Integer observations form a geometric sequence. Interpreting $100(0.8)^t$ at fractional hours assumes a smooth decay between readings; it is an additional modelling choice.

Watch out

A constant loss gives a straight-line model; a constant proportion of the remaining amount gives an exponential model. These descriptions predict different second-hour losses.

Change the reciprocal base once more.

The number e and its graph

The natural exponential has the special property that its tangent gradient always equals its height.

Natural exponential

The natural exponential is $y=\mathrm e^x$ , where $\mathrm e\approx2.718281828$ . It increases through $(0,1)$ , has domain $\mathbb R$ , range $(0,\infty)$ , and horizontal asymptote $y=0$ as $x\to-\infty$ .

The constant $\mathrm e$ is a base, not a variable or a percentage. A tangent records the graph's instantaneous rate of change.

Natural exponential gradient

At input $x$ , the graph $y=\mathrm e^x$ has gradient $\mathrm e^x$ .

Not in the formula sheet: learn it.

For example, chord gradients near zero are $(\mathrm e^{0.01}-1)/0.01\approx1.005$ and $(\mathrm e^{-0.01}-1)/(-0.01)\approx0.995$ . Their approach towards 1 illustrates the property rather than proving it.

Worked example 3: a tangent at height e

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

Plan. Find the point and use its height as the gradient.

\begin{aligned} y&=\mathrm e,\quad m=\mathrm e &&\text{known gradient property}\\ y-\mathrm e&=\mathrm e(x-1) &&\text{point-gradient equation}\\ y&=\mathrm e x &&\text{simplify} \end{aligned}

Why this step? Exact values expose the cancellation of the two constant terms.

Answer. $y=\mathrm e x$ . This tangent passes through the origin, although the exponential itself stays strictly above the horizontal axis.

Check. Substituting $x=1$ gives the contact point $(1,\mathrm e)$ .

Change only the input of the point of contact.

Worked example 4: a steeper tangent

Question. Determine the tangent to $y=\mathrm e^x$ at $x=2$ .

Plan. Use the same property at the new point.

\begin{aligned} y&=\mathrm e^2,\quad m=\mathrm e^2 &&\text{height and gradient}\\ y-\mathrm e^2&=\mathrm e^2(x-2) &&\text{point-gradient equation}\\ y&=\mathrm e^2(x-1) &&\text{collect terms} \end{aligned}

Why this step? The gradient changes with the point; it is not always $\mathrm e$ .

Answer. $y=\mathrm e^2(x-1)$ , with horizontal-axis intercept $(1,0)$ .

Check. At $x=2$ , the line has height $\mathrm e^2$ ; its gradient exceeds the previous tangent's because $\mathrm e^2>\mathrm e$ .

At zero, the same method gives $y=x+1$ , shown beside the curve.

The natural exponential and its tangent at zero

That tangent is a useful local approximation, not an identity for all inputs.

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ME1, ME8, ME9 - Exponential functions and the number e

Positive bases and repeated multiplication

The number e and its graph

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