TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEMS

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Baruch College MTH 2205 Final 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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To view sample exam questions for Baruch MTH 2205 final CLICK HERE.

Baruch MTH 2205 Final Topic 1. Extrema.

There are a few steps behind finding the extrema. Remember, that extrema are at the points where the first derivative is equal to $0$ or does not exist.

Step 1.

Take the first derivative $f\prime(x)$

Step 2.

Set the numerator and denominator of the first derivative to $0$ .

It is key to remember that when it comes down to extrema, it is always about the first derivative. Let’s take a look at the example below from the Sample Final SP17.

MTH 2205, Sample Final SP17, Problem 2

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEM 1

Step 1.

Take the first derivative of $f(x)=\sqrt{4-x^2}$

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

$f(x)=\left(4-x^2\right)^\frac{1}{2}$

$f^\prime\left(x\right)=\frac{1}{2}\ast\left(4-x^2\right)^{-\frac{1}{2}}\ast(-2x)$

$f^\prime\left(x\right)=\frac{1}{2}\ast\frac{1}{\left(4-x^2\right)^\frac{1}{2}}\ast\frac{-2x}{1}$

$f^\prime\left(x\right)=\frac{-2x}{{2\left(4-x^2\right)}^\frac{1}{2}}$

$f^\prime\left(x\right)=\frac{-2x}{2\sqrt{4-x^2}}$

Step 2.

Now simply set both top and bottom to $0$ :

Top:
$-2x=0\rightarrow x=0$

Bottom:
$2\sqrt{4-x^2}=0\rightarrow\ \sqrt{4-x^2}=0\rightarrow4-x^2=0\rightarrow x=\pm2$

We have three answers: $x=0$ , $x=-2$ , $x=2$ .

The answer is $D$ .

Baruch MTH 2205 Final Topic 2. First Derivative Test.

The following easy steps should be followed to find absolute maximum and absolute minimum of a function.

Step 1.

Find the first derivative $f\prime(x)$

Step 2.

Set the first derivative $f^\prime\left(x\right)=0$ and solve for $x$

Step 3.

Reject critical values outside of the range $[a,b]$ set in the problem

Step 4.

Plug the endpoints and remaining critical values into the original function $f(x)$ to find absolute maximum and absolute minimum

We will show how to solve for the absolute maximum and minimum based on a problem from Sample Final FA17.

MTH 2205, Sample Final FA17, Problem 1

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEM 2
We need to find the absolute minimum value of $f\left(x\right)=\frac{2x}{x^2+1}$ on the interval $[0,2]$

Step 1.

To find the first derivative we must use Quotient Rule:

$f\left(x\right)=2x$ and $f\prime\left(x\right)=2$

$g\left(x\right)=x^2+1$ and $g\prime\left(x\right)=2x$

Plug values into the formula:

$f^\prime\left(x\right)=\frac{f^\prime g-fg^\prime}{g^2}=$

$\frac{2\left(x^2+1\right)-2x\ast2x}{\left(x^2+1\right)^2}=$

$\frac{2x^2+2-4x^2}{\left(x^2+1\right)^2}=\frac{-2x^2+2}{\left(x^2+1\right)^2}$

Step 2.

The first derivative must be set to $0$ . It means that the numerator of the fraction must be set to $0$ .

$-2x^2+2=0$

$2x^2=2$

$x^2=1$

$x=\pm1$

Step 3.

Since the given range is $[0,2]$ , we must reject $x=-1$ .

Step 4.

Three points $x=0$ , $x=1$ , and $x=2$ must be plugged into the original function. The smallest value will indicate absolute minimum:

$f\left(0\right)=\frac{2\ast0}{0^2+1}=\frac{0}{1}=0$

$f\left(1\right)=\frac{2\ast1}{1^2+1}=\frac{2}{2}=1$

$f\left(2\right)=\frac{2\ast2}{2^2+1}=\frac{4}{5}$

The smallest value is $f\left(0\right)=0$ , and the answer is $B$ .

Baruch MTH 2205 Final Topic 3. 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 $f\prime\prime(x)$

Step 2.

Set the second derivative $f^\prime\prime\left(x\right)=0$ and solve for $x$

Step 3.

Test intervals by plugging values into $f^\prime\prime\left(x\right)$

MTH 2205, Sample Final SP17, Problem 23

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEM 3
Let’s find the interval when the function $f\left(x\right)=x^4-6x^3+12x^2+4x+4$ is concave downward.

Step 1.

Find the first derivative:

$f\prime(x)=4x^3-18x^2+24x+4$

Take the second derivative:

$f^{\prime\prime}\left(x\right)=12x^2-36x+24$

Step 2.

Set $f^{\prime\prime}\left(x\right)=0$ and solve for $x$ :

$f^{\prime\prime}\left(x\right)=12x^2-36x+24=12\left(x^2-3x+2\right)=0$

$12\left(x-1\right)\left(x-2\right)=0$

$x=1$ and $x=2$

Step 3.

Let’s test $x=1$ and $x=2$ on the number line.

$f^{\prime\prime}\left(0\right)=12\left(0^2-3\ast0+2\right)=12\left(0-0+2\right)=24>0$ , thus the function is concave upward on the interval $(-\infty,1)$

$f^{\prime\prime}\left(1.5\right)=12\left({1.5}^2-3\ast1.5+2\right)=12\left(2.25-4.5+2\right)=-3<0$ , thus the function is concave downward on the interval $(1,2) f^{\prime\prime}\left(3\right)=12\left(3^2-3\ast3+2\right)=12\left(9-9+2\right)=24>0$ , thus the function is concave upward on the interval $(2,\infty)$

The function is concave downward on the interval $(1,2)$ and the answer is $B$ .

Baruch MTH 2205 Final Topic 4. Business Applications.

There are number of rules and formulas for problems involving business applications. Let’s cover one specific case for average revenue, average profit, and average cost.

$Average\ Revenue=\bar{R}(x)=\frac{R\left(x\right)}{x}$

$Average\ Profit=\bar{P}(x)=\frac{P\left(x\right)}{x}$

$Average\ Cost=\bar{C}(x)=\frac{C\left(x\right)}{x}$

Let’s look at the problem that includes the computation of average cost.

MTH 2205, Sample Final FA17, Problem 3

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEM 4

Step 1.

In order to find the average cost of $C\left(x\right)=2x^3-20x^2+100x$ , let’s use the formula above:

$Average\ Cost=\bar{C}(x)=\frac{C\left(x\right)}{x}$

$\bar{C}\left(x\right)=\frac{2x^3-20x^2+100x}{x}=2x^2-20x+100$

Step 2.

Let’s take the first derivative of the average cost:

${\bar{C}}^\prime(x)=4x-20$

Step 3.

By setting the first derivative of the average cost to $0$ we can find the value x that minimizes average cost:

${\bar{C}}^\prime\left(x\right)=4x-20=0$

$4x=20\rightarrow x=5$

The answer is $A$ .

Baruch MTH 2205 Final Topic 5. Linearization.

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEM 5
Linearization problems are, perhaps, the most difficult problems on the Baruch MTH 2205 final. However, they all follow four easy steps. If you know these steps, the problems on linearization become a piece of cake.

The following four steps below will linearize the function:

Step 1.

Take the first derivative $f\prime(x)$

Step 2.

Find $f\prime(x_0)$

Step 3.

Find $f(x_0)$

Step 4.

Find $y=f^\prime\left(x_0\right)\ast\left(x-x_0\right)+\ f(x_0)$

MTH 2205, Sample Final SP17, Problem 25

In Problem 25, $f\left(x\right)=\sqrt{4+x}$ needs to be linearized near $x=0$

Step 1.

The first derivative of $f(x)$ needs to be found. Before we take the first derivative, the original function needs to be rewritten into $f\left(x\right)=\left(4+x\right)^\frac{1}{2}$

$f^\prime\left(x\right)=\frac{1}{2}\ast\left(4+x\right)^{-\frac{1}{2}}=\frac{1}{2}\ast\frac{1}{\left(4+x\right)^\frac{1}{2}}=\frac{1}{2}\ast\frac{1}{\sqrt{4+x}}$

Step 2.

$f^\prime\left(0\right)=\frac{1}{2}\ast\frac{1}{\sqrt{4+0}}=\frac{1}{2}\ast\frac{1}{\sqrt{4+0}}=\frac{1}{2}\ast\frac{1}{2}=\frac{1}{4}$

Step 3.

$f\left(0\right)=\sqrt{4+0}=2$

Step 4.

$y=\frac{1}{4}\ast\left(x-0\right)+\ 2$

$y=\frac{1}{4}x+\ 2$

Step 5.

$f\left(0.12\right)$ needs to be found. Simply plug in $x=0.12$ :

$y\left(0.12\right)=\frac{1}{4}\ast0.12+\ 2=0.03+2=2.03$ .

The answer is $B$ .

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Baruch MTH 2205 Final Topic 6. Inverses.

Inverse problems are, perhaps, one of the easiest types of problems on the Baruch MTH 2205 final. Inverse problems are typically based on pure algebraic manipulations and might have little to do with the actual calculus material. The steps are also very straight-forward.

Step 1.

$y=f(x)$ and $x$ and $y$ values need to be inter-changed.

Step 2.

Solve for $y$

Let’s show how inverse problems can be easily solved with the aforementioned two steps.

MTH 2205, Sample Final FA17, Problem 14

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEM 6
The inverse of $f\left(x\right)=\frac{4x-1}{2x+3}$ should start with the realization that $f\left(x\right)=y$

Re-writing the equation $y=\frac{4x-1}{2x+3}$ always helps with the inverse functions.

Step 1.

Replacing $x$ and $y$ values results in the following:

$x=\frac{4y-1}{2y+3}$

Step 2.

Solve for $y$ :

$\frac{x}{1}=\frac{4y-1}{2y+3}$

$x\ast\left(2y+3\right)=4y-1$

$2xy+3x=4y-1$

$2xy-4y=-3x-1$

$y(2x-4)=-3x-1$

$y=\frac{-3x-1}{2x-4}$

Let’s take a look at the answers. There are no answer choices that match ours. But, we can multiply both top and bottom by $-1$ :

$y=\frac{-1}{-1}\ast\frac{-3x-1}{2x-4}=\frac{3x+1}{4-2x}$

The answer is $D$ .

Baruch MTH 2205 Final Topic 7. Exponential Functions.

There are many types of exponential function problems. We will focus on one of such types in particular.

If the rate of growth is defined, the formula that could be followed is:

$A=P\ast\left(1+r\right)^t$ , where $A$ is the final amount and $P$ is the initial amount.

Let’s attempt the problem directly.

MTH 2205, Sample Final SP17, Problem 30

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEM 7
In this problem, the initial amount is $100,00$ 0 and the final amount if $200,000$ . The annual growth rate is $4.5%$ and we need to find the amount of time it takes to get from the initial amount to the final amount.

Step 1.

$A=P\ast\left(1+r\right)^t$

$200000=100000\ast\left(1+0.045\right)^t$

Step 2.

By plugging the equation in the TI 89 calculator, we can get to the final answer:

The following can be inserted into the calculator:

$solve(200000=100000*(1+0.045)^t,t)$

The answer is: $t=15.7473$

The starting date is $January 1, 2016$ . $15-16$ years later is in between $January 1, 2031$ and $January 1, 2032$ .

So the answer is $B$ .

Baruch MTH 2205 Final Topic 8. Derivative of $\mathbit{e}^\mathbit{x}$ .

Derivatives of exponential functions $e^x$ are very simple. The rule goes like this:

Rule 1.

$f\left(x\right)=e^x\rightarrow f^\prime\left(x\right)=e^x$

Rule 2.

$f\left(x\right)=e^u\rightarrow f^\prime\left(x\right)=e^u\ast u\prime$

Let’s approach the problem where the $u-substitution$ is necessary.

MTH 2205, Sample Final SP17, Problem 12

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEM 8
The first derivative $f^\prime\left(x\right)$ needs to be found for the following function $f\left(x\right)=e^\frac{2}{x}$

Step 1.

Let’s use Rule 2 to find $f^\prime\left(x\right)$ :

$f\left(x\right)=e^\frac{2}{x}$

$f^\prime\left(x\right)=e^\frac{2}{x}\ast\left(\frac{2}{x}\right)^\prime=e^\frac{2}{x}\ast\left(2x^{-1}\right)^\prime=e^\frac{2}{x}\ast-2x^{-2}=-\frac{2}{x^2}e^\frac{2}{x}$

The correct answer is $D$ .

Baruch MTH 2205 Final Topic 9. Logarithms.

Logarithms have the following rules:

Rule 1.

$\log{a}+\log{b}=\log{ab}$

Rule 2.

$\log{a}-\log{b}=\log{\frac{a}{b}}$

Rule 3.

${\log{a}}^n=n\log{a}$

Rule 4.

$\log_a{b}=c\rightarrow a^c=b$

MTH 2205, Sample Final SP17, Problem 18

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEM 9
Problem 18 from Sample File SP17 is a perfect example of a logarithmic problem $\log_2{x}+\log_2{(x+2)}=3$

Step 1.

Let’s combine logarithms together:

$\log_2{x(x+2)}=3$

$\log_2{(x^2+2x)}=3$

Step 2.

Use Rule 4 to create a proper equation and solve for $x$ :

$x^2+2x=2^3$

$x^2+2x=8$

$x^2+2x-8=0$

$x=2 and x=-4$

Step 3.

Final step is to double check both answers and see which answers must be rejected. Remember that the expression inside the logarithm must be strictly greater than $0$ .

$x=2\rightarrow$$ \log_2{2}+\log_2{(2+2)}=1+2=3x=-4\rightarrow$

$\log_2{-4}+\log_2{(-4+2)}=\emptyset$ , since both logarithms have values below $0$ .

The only answer is $x=2$ .

The answer is $B$ .

Baruch MTH 2205 Final Topic 10. Derivative of Logarithms.

On top of the rules from previous section, we should add two more rules. Let’s list all of these rules below:

Rule 1.

$\log{a}+\log{b}=\log{ab}$

Rule 2.

$\log{a}-\log{b}=\log{\frac{a}{b}}$

Rule 3.

${\log{a}}^n=n\log{a}$

Rule 4.

$\log_a{b}=c\rightarrow a^c=b$

Rule 5.

$f\left(x\right)=\ln{x}\rightarrow f\prime\left(x\right)=\frac{1}{x}$

Rule 6.

$f\left(x\right)=\ln{u}\rightarrow f\prime\left(x\right)=\frac{u\prime}{u}$

MTH 2205, Sample Final FA17, Problem 8

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEM 10
Problem 8 from the Sample Final FA17 is a great example of how rules of logarithm can help find the first derivative of $y=\ln{\left(\frac{\sqrt{x-5}}{4x^3+1}\right)}$

Step 1.

Simplify the complicated logarithm:

$y=\ln{\left(\frac{\sqrt{x-5}}{4x^3+1}\right)}=$

$\ln{\sqrt{x-5}}-\ln{\left(4x^3+1\right)}=$

$\ln{\left(x-5\right)^\frac{1}{2}}-\ln{\left(4x^3+1\right)}=$

$\frac{1}{2}\ln{(x-5)}-\ln{\left(4x^3+1\right)}$

Step 2.

Take the first derivative:

$y=\frac{1}{2}\ln{(x-5)}-\ln{\left(4x^3+1\right)}$

$y^\prime=\frac{1}{2}\ast\frac{1}{x-5}-\frac{12x^2}{4x^3+1}=\frac{1}{2(x-5)}-\frac{12x^2}{4x^3+1}$

The answer is $B$ .

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Baruch College MTH 2205 Final Overview

Baruch MTH 2205 Final Topic 1. Extrema.

Step 1.

Step 2.

MTH 2205, Sample Final SP17, Problem 2

Step 1.

Step 2.

Baruch MTH 2205 Final Topic 2. First Derivative Test.

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MTH 2205, Sample Final FA17, Problem 1

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Baruch MTH 2205 Final Topic 3. Concavity.

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Step 3.

MTH 2205, Sample Final SP17, Problem 23

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Baruch MTH 2205 Final Topic 4. Business Applications.

MTH 2205, Sample Final FA17, Problem 3

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Baruch MTH 2205 Final Topic 5. Linearization.

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MTH 2205, Sample Final SP17, Problem 25

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Baruch MTH 2205 Final Topic 6. Inverses.

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MTH 2205, Sample Final FA17, Problem 14

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Baruch MTH 2205 Final Topic 7. Exponential Functions.

MTH 2205, Sample Final SP17, Problem 30

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Baruch MTH 2205 Final Topic 8. Derivative of .

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Rule 2.

MTH 2205, Sample Final SP17, Problem 12

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Baruch MTH 2205 Final Topic 9. Logarithms.

Rule 1.

Rule 2.

Rule 3.

Rule 4.

MTH 2205, Sample Final SP17, Problem 18

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Baruch MTH 2205 Final Topic 10. Derivative of Logarithms.

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Rule 2.

Rule 3.

Rule 4.

Rule 5.

Rule 6.

MTH 2205, Sample Final FA17, Problem 8

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Step 2.

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