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Preview of calculus: part 3

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This is a part of Calculus 1: course.

Exponential Function

Where does it come from?

Remember this simple algebra:

Addition: \( 2 + 2 + 2 = 2 \cdot 3 \rightarrow \) Multiplication

That's how multiplication was "invented" -- as repeated addition. What about repeated multiplication?

Multiplication: \( 2 \cdot 2 \cdot 2 = 2^{3} \rightarrow \) Exponent/Power

So, exponent is a repeated multiplication first. Here we have:

$$ \underset{\textrm{base}}{2}^{\overset{\textrm{exponent}}{3}} $$

With this, we can easily come up with a few formulas...

Algebraic Properties of Exponent

  1. \( a^{m} \cdot a^{n} = a^{m + n} \)
    as in
    $$\begin{aligned}2^{3} \cdot 2^{2} & = (2 \cdot 2 \cdot 2) \cdot (2 \cdot 2 ) \\& = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2& = 2^{5}\end{aligned}$$
    But in addition we also get this:
    $$ 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{\frac{1}{2} + \frac{1}{2}} = 2^{1} $$
    Similar to a rule for multiplication
    $$a \cdot m + a \cdot n = a ( m + n ) $$
  2. \( \dfrac{a^{m}}{a^{n}} = a^{m-n} \)
    (just canceling) as in:
    $$\frac{2^{4}}{2^{2}} = \frac{2 \cdot 2 \cdot 2 \cdot 2}{2 \cdot 2} = 2 \cdot 2 = 2^{2} $$
    Similar to a rule for multiplication:
    $$ a \cdot m - a \cdot n = a ( m - n ) $$
  3. \( (a^{m})^{n} = a^{mn} = (a^{n})^{m} \)
    Exponents.png
    as in
    $$ \begin{aligned}(2^{3})^{2} & = 2^{3} \cdot 2^{3} \\& = ( 2 \cdot 2 \cdot 2 ) ( 2 \cdot 2 \cdot 2 ) \\& = 2^{6}\end{aligned}$$
    Similar to a rule for multiplication:'
    $$ (am)^{n} = a(mn) $$
  4. \( (ab)^{n} = a^{n} b^{n} \) - Distributive property
    Similar to a rule for multiplication:
    \( (a + b)n = an + bn \)

Exponent as a function

We want a function defined for all $x$, i.e., domain = \( (-\infty, \infty) \).

So far, \(a^{x} \) is defined only if \( x \) is a positive integer (and $a>0$).

Let's try to fill in the blanks …

What about \( x = \frac{1}{2} \)?

2tox.png

$$ 2^{\frac{1}{2}} = ? $$ We interpret it as \( \sqrt{2} \), which is a number \( u \) such that $$ u^{2} = 2 $$ Similarly for \( x = \dfrac{1}{3} \), $$ 2^{\frac{1}{3}} = \sqrt[3]{2} $$

More generally, $$ 2^{\frac{p}{q}} = \sqrt[q]{2^{p}} $$

Here \( \dfrac{p}{q} \) is a rational number, i.e., \(p, q \) are integers with \( p \neq 0 \).

This is the interpretation of rational exponents.

RationalExponents.png

There are still gaps!

Indeed irrational $x$'s are missing:

$$x= \sqrt{2}, \sqrt{3}, \pi $$

What happens is roughly that we define the irrational exponents by "approximating" them with rational numbers.

Properties of the exponential function

Let's collect some facts about the graph of this function.

RationalNumbers.png

Range = \( (0, \infty) \).

The \( y \)-intercept is \( (0, 1) \)

No \( x \)-intercepts.

Indeed term "exponential growth" makes sense: \( a ^{x} \) is increasing for \( a > 1 \).

ExponentialGrowth.png

\( y = 0 \) is a horizontal asymptote (later).

Review Exercise. Question: Given functions \( f(x) = \dfrac{x}{1 + x} \), \( g(x) = \sin 2x \), find their compositions, and the domains.

\( f \circ g \) means

CompositionReview.png

Here the output of \( g \) is the input of \( f \).

Substitute: \( y \) into \( f(y) \). $$\begin{aligned} (f \circ g )( x ) & = f(g(x)) = f(y) \\ & = \dfrac{y}{1 + y }, \; \textrm{ use: } y = \sin 2x \\ & = \dfrac{\sin 2x}{1 + \sin 2x} \end{aligned} $$ To find the domain:

Solve: \( 1 + \sin 2x = 0 \) $$ \begin{aligned} \sin 2x &= -1 \\ \Rightarrow 2x & = -\frac{\pi}{2} + 2\pi k \end{aligned} $$ where \( k \) is an integer. So $$ x = -\dfrac{\pi}{4} + \pi k $$ or \( (-\dfrac{\pi}{2}, -\dfrac{\pi}{2} + 2\pi, -\dfrac{\pi}{2} + 4\pi, \ldots ) \)

Answer: The domain is all real numbers except for $$ -\dfrac{\pi}{4} + k\pi $$ where \( k = 0, \pm 1, \pm 2, \ldots \)

  1. \( f \circ f \)

Ff1.png Let $$ f: y = \dfrac{x}{1 + x} $$ and $$ f: z = \dfrac{y}{1 + y} $$ we substitute \( y \) into \( z \). $$ \begin{aligned} z & = \frac{\dfrac{x}{1 + x}}{1 + \dfrac{x}{1+x}} \\ & = \frac{\dfrac{x}{1 + x}}{\dfrac{1+x}{1+x} + \dfrac{x}{1 + x}} \\ & = \frac{\dfrac{x}{1 + x}}{\dfrac{1 + 2x}{1 + x}} \\ & = \dfrac{x}{1 + x} \div \dfrac{1 + 2x}{1 + x} \\ & = \dfrac{x}{1 + 2 x} \end{aligned} $$

Domain: $$ \begin{aligned} x + 1 & \neq 0 \rightarrow x \neq -1, \\ 1 + \frac{x}{1 + x} & \neq 0, \\ 1 + x + x & \neq 0 \\ 1 + 2x & \neq 0 \\ x & \neq -\frac{1}{2} \end{aligned} $$

Answer: All real numbers except \( -1, -\frac{1}{2} \).

So these are two different functions!

Lesson: Don't divide if you are unsure the divisor \( \neq 0 \)

Monotonicity:

$$ \begin{aligned} a^{x} & \textrm{ is increasing if } a > 1 \\ a^{x} & \textrm{ is decreasing if } a < 1 \end{aligned} $$ for \( a > 0 \) and \( a \neq 1 \).

There is an exponential function for each base \( a > 0 \).

Example:

Plot \( h(x) = 2^{3x - 1} \), based on \( f(x) = 2^{x} \).

Operations: \( 3x \) \( -1 \) \( f \)
\( y \) Shrink Horizontally Horizontal Shift to Right \( 3 x \)

Further:

\( 2^{3x} \) looks familiar. It looks like one of the exponential functions.

PlotExercise.png

Compare: \( 2^{3x} \) and \( a^{x} \).

By Property 3, \( 2^{3x} = (2^{3})^{x} = 8^{x} \), it is an exponential function. The one with base 8.

So, turns out \( 8^{x} \) is \( 2^{x} \) shrunk horizontally by \( 3x \).

Example:

Plot \( y = \left(\frac{1}{2}\right)^{x} \), related to \( y = 2^{x} \).

Observe: $$ \left(\frac{1}{2}\right)^{x} = \frac{1}{2^{x}} = 2^{-x} $$

So

\(\left(\frac{1}{2}\right)^{x} \) is \( 2^{x} \) flipped about the \( y \)-axis.

Halfx.png

Lesson: We can get all exponential functions by stretching and flipping one of them.

So, we only need one exponential function!

Which one? \( e^{x} \). It's the natural base exponent. It has a special property: the tangent line at the y-intercept has 45 degree slope, below:

Exponential.png

Exercise:

Given graphs \( f, g \), find graph \( f \circ g \).

Fggraph.png

$$ f \circ g = f(g(x)) $$ for \( x = -5, -4, -3, \ldots \)

Point-by-point:

$$ \begin{aligned} f(g(-5)) & = f(0) = -4 \textrm{ or } (f \circ g)(-5) = -4 \\ f(g(-4)) & = f(1) = -3 \\ f(g(-3)) & = f(1) = -3 \\ \end{aligned} $$

Rewritten:

\( x \) \( g(x) \) \( y \) \( f(y) \) \( z \)
-5   0   -4
-4   1   -3
-3   2   -3

Graphical Solution:

GraphicalSolution.png

Note: \( f \circ g \) requires: \( \textrm{Domain of } f \supset \textrm{ Range of } g \). i.e. all outputs of \( g \) are in the domain of \( f \).

Applications

1. Bacteria double in number everyday.

10 days: \( 10 \cdot 2^{10} = 10.1024 = 10,240 \)

2. Compound Interest, 10% annual, compounded yearly. $$ \begin{aligned}\$ 1000, 1000 \cdot 1.10 & = 1000 + 10\% \textrm { of } 1000 \\ \textrm{begin, after 1 year } & = \textrm{ principal } + \textrm{ interest} \\ & = 1000 + 1000 \cdot .10 \\ & = 1000(1 + 0.1 ) \\ & = 1000 \cdot 1.1 \end{aligned}$$ After \( x \) years: \(1000\cdot 1.1^{x}\), where \( x \) is a positive integer.

Question: How much after 1.5 years?

Interest.png

Use the function: \( 1.1^{x} \) (use calculator)

3. Population Growth: 50 babies per 10,000 of population. $$ \frac{50}{10000} = 0.005 $$ Therefore $$ 1,000,000 \cdot 1.005^{x} $$ What is the multiple? (it's the base of exponent).

Example

Find \( Ca^{x} \) for this graph.

PopulationGrowth.png

Means: Find \( C \) and \( a \).

Use two points to write two equations from \( f(x) = C a^{x} \): $$ (0, 2) \rightarrow x = 0, y = 2 \rightarrow Ca^{0} = 2 \rightarrow C = 2 $$ $$ (2, 2/9) \rightarrow x = 2, y = \frac{2}{9} \rightarrow 2a^{2} = \dfrac{2}{9} \rightarrow a^{2} = \dfrac{1}{9} \rightarrow a = 3. $$

Answer: \( 2 \left( \dfrac{1}{3} \right)^{x}\)

4. City loses 7% of population every year - Exponential Decline.

\( \overbrace{( \underbrace{\underbrace{(1,000 \cdot 0.93 )}_{\textrm{after 1 year}} 0.93}_{\textrm{after 2 years}} ) 0.93}^{\textrm{after 3 years}} \)

Here the multiple is < 1, hence the decline.

ExponentialDecline.png

More New Functions From Old

We know how new algebraic operations may appear as "old" operations are repeated, for example.

Addition : $$ 2 + 2 + 2 = 2 \cdot 3. $$ This leads to a new operation: Multiplication.

Multiplication : $$ 2 \cdot 2 \cdot 2 \cdot 2 = 2^{4}. $$ This leads to a new operation: Exponent.

But what about subtraction and division? How do they appear? It's different.

1. I know addition. Problem: I have \( $5 \), how much do I add to have \( $12\)?

Answer: $7. How do I know? Algebra:

Solve $$ 5 + x = 12. $$ This equation leads to a new operation, subtraction: \( x = 12 - 5 \).

Subtraction is the "inverse" of addition.

2. In know multiplication. Problem: I have a few 2-by-4s, I need a table 20 inches wide. How many do I need?

Answer: 5. How Do I know? Algebra:

Solve $$ 4x = 20 .$$ This equation leads to a new operation, division: \( x = \frac{20}{4} \)

Division is the "inverse" of multiplication.

3. I need a square table of area 25 square feet. What is the width of the table?

Algebra: Solve $$ x \cdot x = 25 \rightarrow x^{2} = 25 \rightarrow x = \sqrt{25} .$$ Thus, we have a new operation: Root.

Root is the "inverse" of square.

4. What about \( \sqrt[3]{} \)?

What is the side of a box given a volume of 8 cubic feet?

Solve $$ x^{3} = 8 \; \; \rightarrow x = \sqrt[3]{8}. $$

5. Bacteria doubles every day. How long does it take to go 8-fold?

Start with equation and solve $$ 2^{x} = 8, $$ where \( x \) is the time in days. Then $$ x = \log_{2} 8. $$ The "logarithm" base 2 is the inverse of the exponential function base 2.

Inverses of functions

Definition. Given a function \( y = f(x) \), the inverse of \( f \) is a function $$ x = f^{-1}(y) $$ such that $$ \begin{aligned} ( f \circ f^{-1} )(y) & = y, \textrm{ for all } x \\ ( f^{-1} \circ f )(x) & = x, \textrm{ for all } y \end{aligned} $$

Note: "\( f^{-1} \)" is the name of the new function.

Diagram representation:

InverseFunction.png

InverseFunction2.png

Example:

InverseFunction3.png

Example:

InverseFunction4.png

Example: If we try this, it doesn't quite work. There is a mismatch.

InverseFunction5.png

Conclusion: Not all functions have inverses.

Solution: make it work by restricting the domain.

Old function: \( y = x^{2} \), with domain = \( (-\infty, \infty) \).

It simply doen;t have the inverse!

New Function : \( y = x^{2} \), with dmain \( [0, \infty ) \).

Then it works: this function has the inverse.

What was the problem? We know that

$$ 2^{2} = 4 $$

But we eliminated this possibility

$$ (-2)^{2} = 4 $$

NotAFunction.png

The inverse of this is not a function as there are two outputs for the same input.

Need to require from a function: Two different inputs can't produce the same output. That will guarantee that the inverse exists.

Definition. \(y = f(x) \) is a one-to-one function if

$$ f(x_{1}) \neq f(x_{2}) $$

unless \( x_{1} = x_{2} \).

Example:

\(x^{2} \) is not one-to one but \( x^{3} \) is.

1to1Function.png

What is the difference?

Two points on the graph have the same height above the x-axis! Hence

Horizontal Line Test. A function is one-to-one if and only if every horizontal line has at most one intersection with its graph.

Test.png

Review Exercise. How are \( e^{x}, e^{-x}, 3e^{x} \) graphically related to each other?

Answer: Shift, stretch and flip.

  • \( e^{-x} \) is \( e^{x} \) flipped about the \( y \)-axis.
  • \( 3e^{x} \) is \( e^{x} \) stretched vertically.

Fact. All increasing functions are 1-1. (Decreasing too).

GraphicalReasoning.png

$$ f\uparrow \textrm{ if: } \underbrace{x_{1} < x_{2}}_{\textrm{Two Different Inputs}} \Longrightarrow \underbrace{f(x_{1}) < f(x_{2})}_{f(x_{1}) \neq f(x_{2})} $$

Hence \( f \) is 1-1.

Fact. Periodic Functions are not 1-1.

PeriodicFunctions.png

Q: How many intersections?

A: Infinitely many.

Then, what about $\sin, \cos, \tan$? Not 1-1!

Then they have no inverses?! No. But what about $\arcsin$, i.e., \( (sin)^{-1} \)?

Answer: As above, limit the domain.

Cosine Pick \( [0, 2\pi] \) as the domain to make it 1-1.

Cos.png

Here \( \cos \) is restricted to \( [0, 2\pi] \) and is decreasing, so it is 1-1. Hence it has an inverse: $$ (\cos)^{-1} $$ or $$ \begin{aligned} (\cos)^{-1}(\cos x) & = x \;\; \textrm{for all } x \textrm{ in } [0, 2\pi] \\ (\cos)(\cos^{-1} y) & = y \;\; \textrm{for all } y \textrm{ in } [-1, 1] \end{aligned} $$

What is the domain of \( \cos^{-1} \)?

  • All possible outputs of \( \cos \).
  • = The range of \( \cos \).
  • = \( [-1, 1] \).

Sine

"New" sin is restricted to \( [-\pi, \pi] \).

It is increasing, so it's 1-1 and have an inverse: $$ (\sin)^{-1} $$

Sin.png

What is the domain of \( \sin^{-1} \)?

  • = the range of \( \sin \).
  • = \( [-1, 1] \).

Check: $$ \begin{aligned} y & = \sin x : f \\ x & = (\sin)^{-1} (y) : f^{-1} \end{aligned} $$

Tangent

Similar.

What is the graph of the inverse?

New function from old, like before...

Q: What do we do with the graph of \( f \) to get the graph \( f^{-1} \)?

A: Technically, we don't need to do anything. Sounds strange, but consider this. The graph of $f$ illustrates how $y$ depends on $x$... and how $x$ depends on $y$. But the latter is $f^{-1}$!

So, there is no need for a new graph -- the graph of $f^-1$ is the graph of $f$! The only problem is that the $x-$ and $y-$axes point in the wrong directions. We just need to fix that.

Sine

Plot \( (\sin)^{-1} \): Flip the whole thing, graph and axes -- about the diagonal -- so that x end up at y and vice versa..

Invsin.png

Invsinflipdiagonal.png $$ x = (\sin)^{-1}(y). $$

Cosine

Invcos.png

Exponent

Invexp.png

Example:

Same idea for any function given only by its graph.

Invfunction.png

Start with choosing a few points on the graph, especially with ones easy to find counterparts (intercepts, points on the diagonal). Plot the counterparts and connect into a curve.

Find the graph of \( x = f^{-1}(y) \):

InvFunction2.png

Review Exercise.

Find the inverse of \( y = \frac{e^{x}}{1 + 2e^{x}} \)

$$ f(x) = \ldots $$

Plan: Solve the equation for \(x\). Check that it is 1-1.

$$ \begin{aligned} y ( 1 + 2e^{x}) & = e^{x} \\ y + y2e^{x} & = e^{x} \\ 2ye^{x} - e^{x} & = -y \\ (2y - 1 ) e^{x} & = - y \\ \therefore e^{x} & = -\frac{y}{2y - 1 } \end{aligned} $$

Apply the inverse to both sides of the equation, which is \( \ln \).

$$ \underbrace{\ln e^{x}}_{x} = \ln\left(-\frac{y}{2y - 1}\right) \rightarrow x = \ln\left(-\frac{y}{2y - 1}\right) $$

Domain: $$ \begin{aligned} 2y - 1 & \neq 0 \\ -\frac{y}{2y - 1} & > 0 \end{aligned} $$

Always: \( \ln ? \) and \( e^{?} \) cancel each other.

Know: $$ \left.\begin{aligned} \ln e^{x} & = x \textrm{ for any } x \\ e^{\ln x} & = x \textrm{ for any } x \end{aligned} \right\} \qquad \textrm{canceling.} $$

Logarithm

Problem: Solve the following for \( x \).

$$ y = 2^{x-5} = 3 $$

To "kill" the exponent and get to $x$, apply its inverse -- to both sides of the equation: (\( \log_{2} y \) (same base). Then

$$ \log_{2} (2^{x - 5}) = \log_{2} (3) $$

Use the property of the inverse and "cancel".

$$ \begin{aligned} x - 5 & = \log_{2} 3 \\ x & = \log_{2} 3 + 5 \end{aligned} $$

If \( a^{x} = y \) then \( x = \log _{a} y \), and vice versa. Recall also this:

If

$$ a \cdot a \cdot \ldots a = ax $$

\( x \) times, then

$$ \log_{a} \underbrace{a \cdot a \cdot \ldots a}_{x \textrm{ times}} = x $$

Examples: $$ \begin{alignat}{2} \log_{10} 100 & = 2 & \quad \log_{5} 125 = 3 \\ \log_{2} \frac{1}{2} & = -1 & \quad \log_{2}\sqrt{2} &= \frac{1}{2} \end{alignat} $$

Definition. \( \log_{a} y \) is the power to which you have to raise \( a \) to get \( y \).

$$ \log_{7} 7 = 1 \qquad \log_{4} 0 \rightarrow \textrm{undefined} $$

Consider: $$ \begin{aligned} \log_{a} \underbrace{ ( a^{m}\cdot a^{n} ) }_{\textrm{multiplication}} & = \log_{a} (a^{m + n}) \\ & = m + n \\ & = \log_{a} a^{m} \underset{\textrm{addition}}{+} \log_{a} a^{n} \\ \therefore \log_{a} (a^{m}\cdot a^{n}) & = \log_{a} a^{m} + \log_{a} a^{n} \end{aligned} $$

Properties of logarithm

  1. $$ \log_{a}(xy) = \log_{a} x + \log_{a} y $$
  2. $$ \log_{a} \left(\frac{x}{y}\right) = \log_{a} x - \log_{a} y $$
  3. $$ \log_{a} (x^{y}) = y \log_{a} x $$

Proof: Given \( x = a^{n} \) $$ \begin{aligned} \log_{a} (a^{n})^{y} & = \log_{a} a^{ny} \\ & = yn \\ & = y \log_{a} x \end{aligned} $$

Each property of $ \log $ corresponds to a property of \( \exp \).

Continue to Limits: part 1.