## Baruch College MTH 2207 – Final Test SP17 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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## Problem 1

### Step 1.

Take the first derivative of

### Step 2.

Remember that

It means that

### Step 3.

The differential of or needs to be found.

Since is changing from to , the and .

Letâ€™s plug in these values:

The answer is .

## Problem 2

Remember that the function has critical numbers when its first derivative is equal to or does not exist.

### Step 1.

Take the first derivative of

It is highly recommended to rewrite the function before taking the derivative:

### Step 2.

Now simply set both top and bottom to :

Top:

Bottom:

We have three answers: , , . The answer is .

## Problem 3

Letâ€™s analyze both conditions presented to us:

and means that the graph of the original function will be always increasing

means that the graph of the original function will be following the concave down concavity. Concave up concavity has a and concave down concavity has the .

The only multiple choice that is both increasing and has a is option , which is the correct answer.

## 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 .

## Problem 5

### 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 .

## Problem 6

The rule of thumb is to select the most complicated-looking expression as u when applying u-substitution method.

### 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 and into the equation

### Step 3.

We have to plug in the value for back

The answer is .

## Problem 7

In related rates problems you must take the derivative of both sides.

### Step 1.

Letâ€™s take the derivative of both sides first:

### Step 2.

Plug in the values for and :

The answer is .

## Problem 8

The average value of on the closed interval can be found by the same formula:

### Step 1.

Letâ€™s first rewrite the equation under the integral and find the integral:

### Step 2.

Simply plug the values into the Average Value function:

The answer is .

## Problem 9

### Step 1.

Take the first derivative of

### Step 2.

Letâ€™s set up the first derivative to , plug in and solve for :

The answer is .

## Problem 10

Keep in mind that is the anti-derivative of

### Step 1.

Take the integral of

### Step 2.

Plug in and solve for :

### Step 3.

Plug in for the integral computed above:

, and the answer is .