Category Archives: MAC1105

Example:

Evaluate (g/f)(-3) when g(x)=x^2-5 and f(x)=sqrt{x+3}

Solution:

Evaluate the function
Rewrite the expression
Replace x with -3 in each function	$(g/f)(-3)={(-3)^2-5}/{sqrt{-3+3}}$
Simplify	$(g/f)(-3)={(-3)^2-5}/{sqrt{-3+3}}$
Division by zero is undefined	is undefined

(g/f)(-3) is undefined.

3.5 Combination of Functions; Composition of Functions, 6.1 Composite Functions

Composition of Functions

November 17, 2017 admin

Example:

Find (f circ g)(x) when

f(x)=8x-9 and g(x)={7}/{x+9}

Solution:

The definition of composition
Replace each x in f with g(x)
Simplify and write a single fraction by finding a common denominator
Simplify by using the distributive property and combine like terms.

(f circ g)(x)=-{9x+25}/{x+9}

5.5 Applications of Exponential and Logarithmic Functions, 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

Application: Exponential Growth and Decay (Half-life)

October 28, 2017 admin

Example: The half life of radium is 1690 years. If 50 grams are present now, how much will be present in 630 years?

Solution: There is a two part process to this problem. Part 1: Use some of the information to find the decay rate of radium. Part 2: Answer the question using the rest of the given information.

Part 1: Find the decay rate of radium.

Since we are using an exponential model for this problem we should be clear on the parts of the exponential decay model.

Exponential Decay Model

$A=A _0 e^{kt}$

A _0 is the initial amount

is the decay rate

is the time

is the amount after t time has passed

Since radium has a half life of 1690 years, we know that after 1690 years there will be half of the initial amount of radium left. This allows me to establish a relationship between the initial amount and the amount after 1690. The amount after 1690 year is half of the initial amount. A=1 /2 A _0 when the t=1690 .

Substitute these values into the exponential decay formula and solve for k.

Exponential Decay Formula	$A=A _0 e^{kt}$
Make a substitution for A and t since it is known that the half-life is 1690 years and	$A=A _0 e^{kt}$ $1 /2 A _0=A _0 e^{k(1690)}$
Solve for the decay rate k:Start by dividing both sides by the coefficient to isolate the exponential factor	$1 /2 A _0=A _0 e^{k(1690)}$ ${1 /2 A _0}/A_0={A _0 e^{k(1690)}}/A_0$ $1 /2= e^{k(1690)}$
Solve for the decay rate k:Take the natural log of both sides to get k out of the exponent	$1 /2= e^{k(1690)}$ $ln(1 /2)= ln(e^{k(1690)})$
Solve for the decay rate k:Use the power rule for logarithms to get k out of the exponent	$ln(1 /2)= ln(e^{k(1690)})$
Solve for the decay rate k:Simplify ln e = 1
Solve for the decay rate k:Solve for k by dividing by 1690 on both sides

Using only the information about radium having a half-life of 1690 years I have found the decay rate for radium.

k={ln(1 /2)}/1690
k approx -0.000410

Note: Although I have put an approximation for k here, try not to round until the very last step.

Part 2: Answer the question using the rest of the given information.

Given information: If 50 grams are present now, how much will be present in 630 years?

With this information I can identify the initial amount of radium as 50 grams and the time to be 630 years. Symbolically that is A _0=50 when the t=630 . Substitute the given information and the decay rate k found in part 1 to the exponential decay formula.

Exponential Decay Formula	$A=A _0 e^{kt}$
Make a substitution for initial amount , time (t), and the decay rate (k). , , and	$A=A _0 e^{kt}$ $A=50 e^{630{ln(1 /2)}/1690}$
Type the expression into your calculator and round to the thousandth place.	$A=50 e^{630{ln(1 /2)}/1690}$

A approx 38.615

38.615 grams will be present 630 years later is 50 grams are present initially.

Here is a video example that is similar to the above example and it shows how to enter information in the calculator.

1.5 Applications of Quadratic Equations, Applications of Quadratic Equations

Application of Quadratic Equation: Translation

October 20, 2017 admin

Example: The product of some negative number and 5 less than twice that number is 273. Find the number.

Solution: Translate the statement into an equation and then solve the equation.

“The product of some negative number…”

Let x be some negative number and multiply that number by the expression that comes next in the statement.

x (expression that comes next)

“…and 5 less than twice that number…”

5 less than means take 5 away from what follows. Twice that number refers to 2 times the negative number that was described before. (2x-5)

x (2x-5)

“…is 273”

This translates to equals 273.

x (2x-5)=273

Solve the equation.

The translated equation
Use the distributed property to rewrite the equation (recognize that the equation is a quadratic equation)
Choose the method for solving the quadratic equation (factoring, square root method, completing the square or quadratic formula) I will demonstrate factoring so I need to move all terms to the same side to have zero on one side.
Factor
Use the zero product property and set each factor equal to zero and solve.

Since the statement says that the number is negative, the number is {-21/2} .

3.2 Properties of a Function's Graph, 3.3 Properties of Functions

Using the Min and Max Feature in the Calculator

September 10, 2017 admin

5.5 Applications of Exponential and Logarithmic Functions, 6.7 Financial Models

Application of Exponential Functions: Finding the Interest Rate

August 4, 2017 admin

Example:

What is the interest rate necessary for an investment to quadruple after 7 year of continuous compound interest?

Solution:

Since this question involve continuous compound interest, we will use the associated formula.

$A=Pe^{rt}$

We are given that the invest quadruples in 7 years. This tells me that when t=7 that A will be 4 times P. I can write that in symbols A=4P.

Substitute these values into the continuous compound formula and solve for the interest rate.

Continuous compound formula	$A=Pe^{rt}$
Substitute the values of t and A into the formula	$A=Pe^{rt}$ $4P=Pe^{r*7}$ $4P=Pe^{7r}$
Solve for r by dividing both sides by P and simplifying	$4P=Pe^{7r}$ ${4P}/P={Pe^{7r}}/P$ $4=e^{7r}$
Solve for r by taking the log of both sides.	$4=e^{7r}$ $ln 4=ln e^{7r}$
Solve for r by using the power rule and simplifying	$ln 4=ln e^{7r}$
Solve for r by dividing both sides by 7 and simplifying
Find the value in the calculator
Write the answer as a percentage rounded to two decimal places

5.4 Exponential and Logarithmic Equations, 6.6 Logarithmic and Exponential Equations, Exponential Equations

Solve an exponential equation: Take the log of both sides

August 3, 2017 admin

Example:

$(1.41)^x = (sqrt{2})^{1-4x}$

Solution:

The exponential equation	$(1.41)^x = (sqrt{2})^{1-4x}$
Since the bases cannot be easily written the same use the method of taking the log of both sides	$(1.41)^x = (sqrt{2})^{1-4x}$ $ln (1.41)^x = ln (sqrt{2})^{1-4x}$
Use the power rule for logarithms	$ln (1.41)^x = ln (sqrt{2})^{1-4x}$
Use the distributive law
Collect the terms with x to one side and collect the terms without x on the other side
Factor the common x
Solve for x by dividing both sides by the factor in the parenthesis and simplify
The solution

When you type this into a calculator be sure to use parenthesis around the numerator and around the denominator. Here is an example of how you might enter it.

(ln (sqrt{2}))/(ln(1.41)+4 ln(sqrt{2}))

Here is a youtube video with a similar example.

5.3 Properties of Logarithms, 5.4 Exponential and Logarithmic Equations, 6.6 Logarithmic and Exponential Equations, Logarithmic Equations

Solve the Logarithmic Equation by the one to one property

August 3, 2017 admin

Example:

2 log_3(7-x)-log_3 2=log_3 18

Solution:

The logarithmic equation
Use the power rule and the quotient rule to condense to a single logarithm
Since both sides of the equation have the same log base the expressions inside the logarithms must be equal
Clear the denominator by multiplying by 2 on both sides and simplifying
Get rid of the square by square rooting both sides and simplifying	$sqrt{(7-x)^2}= sqrt{36}$
Get x by itself by subtracting 7 on both sides
Get x by itself by dividing both sides by negative 1	or or
Check x=13	Log of a negative is undefined. Exclude this solution.
Check x=1	Keep this solution.

The solution to the equation is x=1.

Here is a youtube video that is similar.

5.4 Exponential and Logarithmic Equations, 6.6 Logarithmic and Exponential Equations

Solve an Exponential Equation: Take the log of both sides

August 1, 2017 admin

Problem: Solve the exponential equation.

$16^{3x-3}=3^{x-3}$

Solution:

The exponential equation	$16^{3x-3}=3^{x-3}$
Since the bases cannot be easily written the same, use the method of taking the log of both sides	$ln (16^{3x-3})=ln (3^{x-3})$
Use the power rule for logarithms.
Use the distributive property.
Collect the terms with x to one side and collect the terms without x on the other side.
Factor the common x.
Solve for x by dividing both sides by the factor in the parenthesis and simplify.
The solution

When you type this into a calculator be sure to use parenthesis around the numerator and around the denominator. Here is an example of how you might enter it.

(3ln(16)-3ln(3))/(3ln(16)-ln(3))

5.1 Exponential Functions, 6.6 Logarithmic and Exponential Equations, Exponential Equations

Solving an Exponential Equation: Relating the Bases

July 27, 2017 admin

Example: Solve the exponential equation.

$4^{x-9}=1/1024$

Solution:

The exponential equation	$4^{x-9}=1/1024$
Try to write both sides of the equation with the same base. Try 4 since there is a base of 4 on the left	$4^{x-9}=1/4^5$
Using a property of negative exponents move the base to the numerator	$4^{x-9}=4^{-5}$
Now that that the bases are the same the exponents must be equal	$4^{x-9}=4^{-5}$
Solve for x