Review of Exponential Functions Base e Revision Notes for HSC SSCE Mathematics Advanced

Review of Exponential Functions Base e

Introduction

This section reviews exponential functions with base $e$ from Year 11, with two new topics added:

Dilations of exponential functions
Exponential and logarithmic equations reducible to quadratics (covered in Section 5F)

The exponential function $y = e^x$ has special properties that make it fundamental in mathematics and science. This note focuses on understanding $e^x$ , its transformations, and applications to tangent and normal lines.

The number e and the function y = e^x

What is e?

The number e is an irrational number, approximately equal to 2.7183. It is defined as the unique base for which the exponential function is its own derivative. This means:

$\frac{d}{dx}e^x = e^x$

In other words, at every point on the curve $y = e^x$ , the gradient of the tangent line equals the y-coordinate.

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The Fundamental Property: "Gradient Equals Height"

This property is what makes $e$ special among all exponential functions. At any point $(x, e^x)$ on the curve, the slope of the tangent line is exactly equal to the y-coordinate $e^x$ .

This is the defining characteristic of the natural exponential function.

The number $e$ plays a similar role in exponential functions as $\pi$ plays in trigonometric functions - it's a fundamental constant that appears throughout mathematics.

Key properties of y = e^x

The exponential function $y = e^x$ has several important characteristics:

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Derivative property:

The function is its own derivative: $\frac{d}{dx}e^x = e^x$
At the y-intercept (where $x = 0$ ), the gradient is exactly 1
At each point, gradient equals height

Domain and range:

Domain: All real numbers ( $x \in \mathbb{R}$ )
Range: $y > 0$ (the function is always positive)

Asymptote:

The line y = 0 (the x-axis) is a horizontal asymptote
As $x \to -\infty$ , the function approaches 0 but never reaches it

Other properties:

The function is one-to-one, meaning its inverse relation is also a function
The y-intercept is at (0, 1) because $e^0 = 1$
The second derivative is also $e^x$ : $\frac{d^2}{dx^2}e^x = e^x$
The function is always concave up and increasing at an increasing rate

A sketch of $y = e^x$ shows an exponential curve passing through $(0, 1)$ and $(1, e)$ , rising steeply as $x$ increases and approaching the x-axis as $x$ decreases.

Inverse function identities

The natural logarithm $\log_e x$ (also written as $\ln x$ ) is the inverse function of $e^x$ . Two important identities connect these functions:

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Key Inverse Function Relationships:

$\log_e e^x = x \text{ for all real } x$

$e^{\log_e x} = x \text{ for } x > 0$

These identities are useful when simplifying expressions and solving equations.

Using the calculator

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Calculator Keys for Exponential and Logarithmic Functions:

On your calculator:

The button ln represents $\log_e x$ (natural logarithm)
The button log represents $\log_{10} x$ (common logarithm)
The function $e^x$ is usually accessed using shift followed by ln, or a similar key sequence

Transformations of y = e^x

We can create new exponential functions by applying transformations to the basic function $y = e^x$ . The main transformations are reflections and translations.

Reflection in the y-axis: y = e^-x

The function $y = e^{-x}$ is obtained by reflecting $y = e^x$ in the y-axis.

Properties of y = e^-x:

Reflects the original curve horizontally
y-intercept: $(0, 1)$ (same as $y = e^x$ )
Horizontal asymptote: $y = 0$
Range: $y > 0$
This function represents exponential decay rather than growth

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A reflection in the y-axis can also be thought of as a horizontal dilation with factor $-1$ .

Vertical translation: y = e^x + k

Adding a constant $k$ to $e^x$ shifts the entire graph vertically.

For y = e^x + 3:

Shift the graph of $y = e^x$ up by 3 units
y-intercept: $(0, 4)$ because $e^0 + 3 = 1 + 3 = 4$
Horizontal asymptote: y = 3 (shifted up from $y = 0$ )
Range: $y > 3$

Horizontal translation: y = e^(x-h)

Replacing $x$ with $(x - h)$ shifts the graph horizontally.

For y = e^(x-2):

Shift the graph of $y = e^x$ right by 2 units
y-intercept: $(0, e^{-2})$ because when $x = 0$ , $y = e^{-2} \approx 0.135$
Horizontal asymptote: $y = 0$ (unchanged)
Range: $y > 0$

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Connection Between Shifts and Dilations:

The function $y = e^{x-2}$ can be rewritten as $y = e^{-2} \cdot e^x$ , which means it's also a vertical dilation of $y = e^x$ with factor $e^{-2} \approx 0.135$ .

This shows that horizontal shifts of exponential functions can sometimes be expressed as vertical dilations!

Dilations of y = e^x

Dilations were introduced in Section 2G and involve stretching or compressing the graph either horizontally or vertically. With exponential functions, dilations have an interesting property: some can be expressed as shifts in the other direction.

Horizontal dilation: y = e^(ax)

Replacing $x$ with $ax$ creates a horizontal dilation with factor $\frac{1}{a}$ .

For y = e^(3x):

Dilate $y = e^x$ horizontally with factor $\frac{1}{3}$
This compresses the graph horizontally (makes it steeper)
The y-intercept remains at $(0, 1)$
When $x = \frac{1}{3}$ , $y = e$ (the curve reaches height $e$ sooner)

Vertical dilation: y = ae^x

Multiplying $e^x$ by a constant $a$ creates a vertical dilation with factor $a$ .

For y = 3e^x:

Dilate $y = e^x$ vertically with factor 3
This stretches the graph vertically
y-intercept: $(0, 3)$ because $3e^0 = 3$
When $x = 1$ , $y = 3e$ (three times higher than the original curve)

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Another Connection: Vertical Dilation as Horizontal Shift

The function $y = 3e^x$ can be rewritten as: $y = e^{\log_e 3} \cdot e^x = e^{x + \log_e 3}$

This means it can also be viewed as a horizontal shift left by $\log_e 3$ units.

Key insight about dilations

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Special Property of Exponential Function Dilations:

Some dilations of exponential functions can be expressed in two ways:

$y = e^{ax}$ is a horizontal dilation and cannot be rewritten as a simple shift
$y = ae^x$ is both a vertical dilation with factor $a$ AND a horizontal shift left by $\log_e a$

This special property arises from the index law: $ae^x = e^{\log_e a} \cdot e^x = e^{x + \log_e a}$

Tangents and normals to the exponential function

The derivative property $\frac{d}{dx}e^x = e^x$ makes finding tangent and normal lines straightforward. Let's work through a complete example.

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Worked Example: Tangent and Normal to y = 2e^x at x = 1

Question: Let $A$ be the point on the curve $y = 2e^x$ where $x = 1$ .

a) Find the equations of the tangent and normal at point $A$

b) Show that the tangent passes through the origin, and find where the normal meets the x-axis (point $B$ )

c) Find the area of triangle $AOB$

Solution:

Part (a): Finding point A and the gradient

First, find the y-coordinate of point $A$ by substituting $x = 1$ :

$y = 2e^1 = 2e$

So A = (1, 2e).

Next, find the gradient by differentiating:

$y = 2e^x$

$\frac{dy}{dx} = 2e^x$

At point $A$ where $x = 1$ :

$\frac{dy}{dx} = 2e^1 = 2e$

This confirms the "gradient equals height" property since the gradient is $2e$ , which equals the y-coordinate.

Finding the tangent equation:

Using point-gradient form with $m = 2e$ and point $(1, 2e)$ :

y - y_1 = m(x - x_1) y - 2e = 2e(x - 1) y - 2e = 2ex - 2e y = 2ex

Tangent equation: y = 2ex

Finding the normal equation:

The normal is perpendicular to the tangent, so its gradient is $-\frac{1}{2e}$ (negative reciprocal).

Using point-gradient form:

$y - 2e = -\frac{1}{2e}(x - 1)$

Multiply both sides by $2e$ :

2ey - 4e^2 = -(x - 1) 2ey - 4e^2 = -x + 1 x + 2ey = 4e^2 + 1

Normal equation: x + 2ey = 4e² + 1

Part (b): Intercepts

Tangent through origin:

The tangent equation is $y = 2ex$ . When $x = 0$ , we have $y = 0$ , so the tangent passes through the origin $O = (0, 0)$ . Its y-intercept is zero.

Normal x-intercept:

To find where the normal meets the x-axis (point $B$ ), set $y = 0$ in the normal equation:

x + 2e(0) = 4e^2 + 1 x = 4e^2 + 1

So B = (4e² + 1, 0).

Part (c): Area of triangle AOB

Triangle $AOB$ has:

Base along the x-axis from $O$ to $B$ : length $= 4e^2 + 1$
Height perpendicular to the base: height $= 2e$ (the y-coordinate of $A$ )

Using the triangle area formula:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (4e^2 + 1) \times 2e = e(4e^2 + 1) \text{ square units}

Area = e(4e² + 1) square units

This example demonstrates how the derivative property of $e^x$ simplifies calculations involving tangent and normal lines.

Remember!

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Key Points to Remember:

The defining property: The exponential function $y = e^x$ is its own derivative: $\frac{d}{dx}e^x = e^x$ . At every point, gradient equals height.
Key transformations: Reflecting $y = e^x$ in the y-axis gives $y = e^{-x}$ (exponential decay). Vertical shifts change the asymptote from $y = 0$ to $y = k$ . Horizontal shifts can also be expressed as vertical dilations.
Dilations and shifts: Some vertical dilations of $e^x$ can be rewritten as horizontal shifts. For example, $y = ae^x = e^{x + \log_e a}$ is both a vertical stretch and a horizontal shift.
Inverse function: The natural logarithm $\log_e x$ is the inverse of $e^x$ . Remember: $\log_e e^x = x$ and $e^{\log_e x} = x$ .
Tangent and normal lines: Use the derivative to find the gradient of the tangent. The normal's gradient is the negative reciprocal. Both can be found using point-gradient form: $y - y_1 = m(x - x_1)$ .

Review of Exponential Functions Base e (HSC SSCE Mathematics Advanced): Revision Notes