TABLE OF CONTENTS

## Baruch College MTH 2205 – Final Test SP17 Overview

MTH 2205 is a calculus course in Baruch College. Its prerequisite is MTH 2003 or MTH 2009. Baruch College MTH 2205 Final Test has a calculator section. 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:

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.

Find the points at which right-hand approximations need to be counted.

The boundaries are and

There are rectangles, so the width of each rectangle is:

Since we are looking at the RIGHT-hand approximation, the points are , , and .

### Step 2.

Plug in the values into the following equations

Letâ€™s find the values of individual elements:

The correct 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

The formula for the elasticity of demand is

### Step 1.

Based on the formula above, we need to have values for , , and .

We already know that . Letâ€™s find the value for :

### Step 2.

Letâ€™s find

### Step 3.

Letâ€™s compute , when :

### Step 4.

Elasticity of demand can be found by plugging all the components:

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 .