TABLE OF CONTENTS

## Baruch MTH 2207 Final Overview

MTH 2207 is a calculus course in Baruch College. This course is a combination of MTH 2205 and algebra of matrices. This course is not open to students who completed MTH 2003 or MTH 2009; these students take MTH 2205. Therefore, the students are required to master TI 89 or TI 92 graphical calculator for this course.

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## Baruch MTH 2207 Final Topic 1. Limits and Continuity.

When it comes down to limits and continuity problems, it is important to check the boundaries. Typical steps include plugging in all the border values into function ) and making sure these values are either matching or not. Matching values indicate that there is no discontinuity. Non-matching values indicate that there is a discontinuity.

Letâ€™s take a look at the specific problem below.

## MTH 2207, Sample Final SP17, Problem 25

In the final test SP17 of MTH 2207 the following piece-wise function needs to be examined for continuity

### Step 1.

Letâ€™s plug into every single equation given

There is a clear disconnect at . Thus the function is continuous between and .

The answer is .

## Baruch MTH 2207 Final Topic 2. Asymptotes.

Vertical Asymptotes

The steps for identifying vertical asymptotes are quite straight-forward.

### Step 1.

Simplify the expression

### Step 2.

Set the denominator to .

Horizontal Asymptotes

Select the highest exponent with the coefficient on top and bottom:

### Step 1.

If the exponent of top is greater than the exponent on the bottom -> undefined

### Step 2.

If the exponent of top is equal to the exponent on the bottom -> fraction of coefficients

### Step 3.

If the exponent of top is less than the exponent on the bottom ->

## MTH 2207, Sample Final SP17, Problem 5

Letâ€™s take a look at the function

### Step 1.

Take a look at the highest exponent of the top and the bottom.

The highest exponent of the numerator is .

The highest exponent of the denominator is .

Since the exponent of denominator is higher, the horizontal asymptote is .

### Step 2.

To find the vertical asymptote, simplify the fraction and set the denominator to .

Letâ€™s factor out both top and bottom:

Letâ€™s set the denominator to :

The horizontal asymptote is and the vertical asymptote is .

The answer is .

## Baruch MTH 2207 Final Topic 3. Derivative of Logarithms.

### Rule 1.

### Rule 2.

## MTH 2207, Sample Final SP17, Problem 4

Keep in mind that the confusing term is just the first derivative.

### Step 1.

Take the first derivative of using the Product and Chain rule.

and

and

Plug values into the formula:

The correct answer is .

## Baruch MTH 2207 Final Topic 4. Product and Quotient Rules.

### Product Rule.

### Quotient Rule.

Letâ€™s use these formulas to solve Problem 6 from the Sample Final FA17.

## MTH 2207, Sample Final FA17, Problem 6

### Step 1.

Take the first derivative of using the Product and Chain rule.

and

and

Plug values into the formula:

### Step 2.

Plug in to find

The correct answer is .

## Baruch MTH 2207 Final Topic 5. Chain Rule.

### Chain Rule.

Chain Rule application is very simple to use as long as you do not forget to multiply the end result by the derivative of the inside .

## MTH 2207, Sample Final SP17, Problem 21

### Step 1.

Make the U-substitution for

The answer is .

## Baruch MTH 2207 Final Topic 6. Slope and Linearization.

Linearization problems are, perhaps, the most difficult problems on the Baruch MTH 2207 final. The following four steps will make it much easier to solve such problems.

The following four steps below will linearize the function:

### Step 1.

Take the first derivative

### Step 2.

Find â€“ this is the slope

### Step 3.

Find

### Step 4.

Find

## MTH 2207, Sample Final SP17, Problem 13

Letâ€™s linearize the following function at the point

### Step 1.

### Step 2.

### Step 3.

### Step 4.

The answer is .

## Baruch MTH 2207 Final Topic 7. Concavity.

The following easy steps will give you inflection points and should be followed to find absolute maximum and intervals on which the function is concave upward and concave downward.

### Step 1.

Find the second derivative

### Step 2.

Set the second derivative and solve for

### Step 3.

Test intervals by plugging values into

## MTH 2207, Sample Final FA17, Problem 9

Letâ€™s take a look at Problem 9 from the Sample Final FA17. The second derivative is already found in the problem and is given to us and we are asked to find the intervals on which this function is concave up.

### Step 1.

Letâ€™s factor this function further:

### Step 2.

### Step 3.

Letâ€™s now test these inflection points by plugging values around them:

We can conclude the following:

The function is concave up for the following intervals: .

The answer is .

## Baruch MTH 2207 Final Topic 8. Differentials.

Differential problems include proper computation of and/or .

### Rule 1.

### Rule 2.

## MTH 2207, Sample Final SP17, Problem 1

Differential problems can be simple if you know the rules. Letâ€™s find the differential for .

### Step 1.

Take the first derivative of

### Step 2.

The differential of or needs to be found.

Since is changing from to , the and .

Letâ€™s plug in these values:

The answer is .

## Baruch MTH 2207 Final Topic 9. Implicit Differentiation.

In Implicit Differentiation problems, it is imperative to find . There are two simple rules for implicit differentiation problems.

### Rule 1.

Every time you take the derivative of , follow derivative rules.

### Rule 1.

Every time you take the derivative of , multiply the final result by .

## MTH 2207, Sample Final SP17, Problem 22

Letâ€™s find the slope of the tangent line to the following expression at the point .

### Step 1.

Take the derivative of both sides of the expression:

(

a.

b.

Letâ€™s plug in the values into the following expression:

c.

### Step 2.

Letâ€™s now plug in the value into the first derivative expression:

The answer is .

## Baruch MTH 2207 Final Topic 10. Antiderivatives.

In antiderivative problems, take a look at the expression. If a fraction is given, notice if the top looks more difficult that the bottom. If thatâ€™s the case, make the U-substitution for the bottom.

## MTH 2207, Sample Final SP17, Problem 6

### Step 1.

The more complex-looking expression in the fraction is

Keep in mind that , so we can say the following:

Solving for we get the following:

### Step 2.

Plug in u and dx into the equation

### Step 3.

We have to plug in the value for back

The answer is .