Unit 5, lesson 38

The Natural Exponential Function

Why e is an exact constant and not a rounded decimal, what the graph of the natural exponential looks like with its exact landmarks, and how constant multiples in front and in the exponent model growth and decay.

  • Theory part 75
  • Practice part 76
  • Exponential Functions

Out of all the positive bases, calculus picks one, the number ee, because its exponential function has the simplest rate of change of any. You do not need that reason yet to use it. Everything you know about exponential functions with other bases still applies to exe^x, and the only new habit is to treat ee as an exact number.

Key terms

The number e

Definition The number ee is the exact real constant

e=lim⁡n→∞(1+1n)n=2.718281828…,\begin{aligned} e &= \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \\[0.4em] &= 2.718281828\ldots, \end{aligned}

where nn runs through the positive integers 1,2,3,…1, 2, 3, \dots It is irrational, so its decimal expansion never ends and never repeats. The decimal 2.718282.71828 is an approximation of ee, not its definition. Here we take on trust that this limit exists. The proof, and the property of ee that makes it special, come in calculus.

In plain words The letter ee names one particular number a little less than 2.722.72, in the same way that π\pi names one number a little more than 3.143.14.

Example The values of (1+1n)n\left(1 + \frac{1}{n}\right)^n creep up toward ee.

nn(1+1n)n\left(1 + \frac{1}{n}\right)^n
1122
10102.59374…2.59374\ldots
1001002.70481…2.70481\ldots
100010002.71692…2.71692\ldots
10610^62.71828…2.71828\ldots

The natural exponential and its positive base

Definition The natural exponential function is

f(x)=ex,x∈R.f(x) = e^x, \qquad x \in \mathbb{R}.

Its base ee is positive and not equal to 11, so it belongs to the family axa^x with a>0a > 0, a≠1a \ne 1. Because e>1e > 1, the function is strictly increasing on R\mathbb{R}, and ex>0e^x > 0 for every real xx. Its range is (0,∞)(0, \infty). For all real xx and yy,

e0=1,e−x=1ex,ex+y=exey.\begin{gathered} e^0 = 1, \qquad e^{-x} = \frac{1}{e^x}, \\[0.4em] e^{x + y} = e^x e^y. \end{gathered}

In plain words The graph of exe^x has the same shape as 2x2^x and 3x3^x and lies between them, since 2<e<32 < e < 3. It passes through (0,1)(0, 1) and (1,e)(1, e) and climbs faster and faster to the right.

Example The main landmarks, exact first and rounded second, are

xx−2-2−1-1001122
exact1e2\frac{1}{e^2}1e\frac{1}{e}11eee2e^2
approximate0.1350.1350.3680.368112.7182.7187.3897.389

Exact and approximate values

Definition An exact value is written with symbols that name the number precisely, such as e2e^2, 1e\frac{1}{e} or 3e0.53e^{0.5}. An approximate value is a decimal rounded to some number of places, such as 7.3897.389, written with ≈\approx rather than ==. Rounding intermediate results changes later results, so a careful computation keeps exact values until the last step.

In plain words Keep ee as ee while you work. Round only when you need a number to read off or report.

Example Exactly,

e0.5⋅e1.5=e2≈7.389.e^{0.5} \cdot e^{1.5} = e^{2} \approx 7.389.

If you first round e0.5≈1.65e^{0.5} \approx 1.65 and e1.5≈4.48e^{1.5} \approx 4.48, the product is 7.3927.392, which is already wrong in the third decimal place.

Fixed-factor growth

Definition For any fixed increment Δ\Delta and every real xx,

ex+Δex=eΔ.\frac{e^{x + \Delta}}{e^x} = e^{\Delta}.

The same holds for f(t)=Cektf(t) = C e^{kt} with C≠0C \ne 0, where

f(t+Δ)f(t)=ekΔ.\frac{f(t + \Delta)}{f(t)} = e^{k\Delta}.

The factor depends only on the increment, not on the starting input.

In plain words Over any input step of the same length, an ee-based quantity is multiplied by the same number, exactly as with any other base.

Example An amount A(t)=50e0.2tA(t) = 50e^{0.2t} is multiplied by e0.2⋅5=ee^{0.2 \cdot 5} = e over every 55 units of time.

tt005510101515
A(t)A(t) exact505050e50e50e250e^250e350e^3
A(t)A(t) approximate5050135.91135.91369.45369.451004.281004.28

Exponential decay

Definition The function e−x=(1e)xe^{-x} = \left(\frac{1}{e}\right)^x has base 1e\frac{1}{e}, which lies between 00 and 11, so it is strictly decreasing on R\mathbb{R}. A model

Q(t)=Q0e−kt,Q0>0, k>0,\begin{gathered} Q(t) = Q_0 e^{-kt}, \\[0.4em] Q_0 > 0,\ k > 0, \end{gathered}

describes exponential decay. It starts at Q(0)=Q0Q(0) = Q_0 and keeps the fraction e−kΔe^{-k\Delta} of its value over every input step Δ\Delta.

In plain words A negative coefficient in the exponent turns growth into decay. Each step keeps the same fraction of what was there.

Example For Q(t)=80e−0.5tQ(t) = 80e^{-0.5t}, every 22 units of time multiply the amount by e−1≈0.368e^{-1} \approx 0.368. So Q(2)=80e≈29.43Q(2) = \frac{80}{e} \approx 29.43 and Q(4)=80e2≈10.83Q(4) = \frac{80}{e^2} \approx 10.83.

Horizontal asymptote

Definition As x→−∞x \to -\infty the values exe^x approach 00, and as x→∞x \to \infty the values e−xe^{-x} approach 00. So y=0y = 0 is a horizontal asymptote of both graphs, on one side each, and neither graph reaches it. For f(x)=Cekx+Lf(x) = C e^{kx} + L with C≠0C \ne 0 and k≠0k \ne 0, the horizontal asymptote is y=Ly = L.

In plain words Far to one side, the graph flattens against a horizontal line without touching it. Adding a constant lifts that line.

Example The function g(x)=3−2e−xg(x) = 3 - 2e^{-x} has g(0)=1g(0) = 1. As xx grows, 2e−x2e^{-x} shrinks toward 00, so g(x)g(x) rises toward 33 but stays below it. The line y=3y = 3 is its horizontal asymptote, and the range of gg is (−∞,3)(-\infty, 3).

Scaled exponential

Definition A scaled exponential is a function

f(x)=Cekx,C≠0, k≠0.f(x) = C e^{kx}, \qquad C \ne 0,\ k \ne 0.

Since ekx=(ek)xe^{kx} = \left(e^k\right)^x, it is an exponential function with base eke^k. For k>0k > 0 the base is greater than 11, and for k<0k < 0 it lies between 00 and 11. The graph of y=ekxy = e^{kx} is the graph of y=exy = e^x scaled horizontally by the factor 1∣k∣\frac{1}{\lvert k \rvert} toward or away from the yy-axis, a compression when ∣k∣>1\lvert k \rvert > 1 and a stretch when ∣k∣<1\lvert k \rvert < 1, and also reflected across the yy-axis when k<0k < 0. Conversely, every base a>0a > 0 with a≠1a \ne 1 equals eke^k for exactly one real number k≠0k \ne 0, so every exponential function can be written with base ee. That kk is the natural logarithm of aa, which the natural logarithm lesson introduces.

In plain words The number kk controls how fast things happen. Doubling kk makes the same changes happen in half the input distance.

Example The function y=e2xy = e^{2x} has base e2≈7.389e^2 \approx 7.389. It reaches the height ee at x=12x = \frac{1}{2}, where exe^x needs x=1x = 1. The function y=e−0.5xy = e^{-0.5x} has base e−0.5≈0.607e^{-0.5} \approx 0.607, so it decays, reaching 1e\frac{1}{e} at x=2x = 2.

Check yourself

  1. Simplify to a single exact power of ee.
e3⋅e−5e−1\frac{e^3 \cdot e^{-5}}{e^{-1}}
Show the answer

Adding and subtracting exponents gives 3−5−(−1)=−13 - 5 - (-1) = -1, so the value is e−1=1e≈0.368e^{-1} = \frac{1}{e} \approx 0.368.

  1. For f(t)=12e0.25tf(t) = 12e^{0.25t}, by what factor does ff change from t=4t = 4 to t=8t = 8? Give f(8)f(8) exactly and approximately.
Show the answer

The step is Δ=4\Delta = 4, so the factor is e0.25⋅4=ee^{0.25 \cdot 4} = e. The value is f(8)=12e2≈88.67f(8) = 12e^2 \approx 88.67.

  1. Find the yy-intercept, the horizontal asymptote and the direction of change of g(x)=5+2e−3xg(x) = 5 + 2e^{-3x}.
Show the answer

The intercept is g(0)=5+2=7g(0) = 5 + 2 = 7. As x→∞x \to \infty the term 2e−3x2e^{-3x} shrinks to 00, so the asymptote is y=5y = 5. The base e−3e^{-3} is between 00 and 11 and the coefficient 22 is positive, so gg is decreasing.

How this lesson connects