TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEMS

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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.

To view the syllabus for Baruch College MTH 2207 CLICK HERE

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To view sample final exam questions for Baruch MTH 2207 CLICK HERE.

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 $f(x$ ) 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.

Let’s take a look at the specific problem below.

MTH 2207, Sample Final SP17, Problem 25

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEM 1
In the final test SP17 of MTH 2207 the following piece-wise function needs to be examined for continuity

$f\left(x\right)=\left\{\begin{matrix}x^2+1,\ x<2\\8,\ x=2\\3x-1,\ x>2\\\end{matrix}\right.$

Step 1.

Let’s plug $x=2$ into every single equation given

$f\left(2\right)=\left\{\begin{matrix}2^2+1=5,\ x<2\\8,\ x=2\\3\ast2-1=5,\ x>2\\\end{matrix}\right.$

There is a clear disconnect at $x=2$ . Thus the function is continuous between $(-\infty,2)$ and $(2,\infty)$ .

The answer is $E$ .

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 $0$ .

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 -> $0$

MTH 2207, Sample Final SP17, Problem 5

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEM 2
Let’s take a look at the function $y=\frac{4x-8}{x^2+5x-14}$

Step 1.

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

The highest exponent of the numerator $4x-8$ is $1$ .

The highest exponent of the denominator $x^2+5x-14$ is $2$ .

Since the exponent of denominator is higher, the horizontal asymptote is $y=0$ .

Step 2.

To find the vertical asymptote, simplify the fraction and set the denominator to $0$ .

$y=\frac{4x-8}{x^2+5x-14}$

Let’s factor out both top and bottom:

$y=\frac{4(x-2)}{(x-2)(x+7)}=\frac{4}{x+7}$

Let’s set the denominator to $0$ :

$x+7=0\rightarrow x=-7$

The horizontal asymptote is $y=0$ and the vertical asymptote is $x=-7$ .

The answer is $A$ .

Baruch MTH 2207 Final Topic 3. Derivative of Logarithms.

Rule 1.

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

Rule 2.

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

MTH 2207, Sample Final SP17, Problem 4

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEM 3
Keep in mind that the confusing $\frac{dy}{dx}$ term is just the first derivative.

Step 1.

Take the first derivative of $f\left(x\right)=x\ast\ln{(x^2})$ using the Product and Chain rule.

$f=x$ and $f\prime=1$

$g=\ln\funcapply(x^2)$ and $g^\prime=\frac{2x}{x^2}=\frac{2}{x}$

Plug values into the formula:

$f^\prime\left(x\right)=f^\prime g+fg^\prime=$

$x\ast\frac{2}{x}+1\ast\ln{\left(x^2\right)}=2+\ln\funcapply(x^2)$

The correct answer is $A$ .

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

Product Rule.

$\left(f\ast g\right)^\prime=f^\prime g+fg\prime$

Quotient Rule.

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

Let’s use these formulas to solve Problem 6 from the Sample Final FA17.

MTH 2207, Sample Final FA17, Problem 6

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEM 4

Step 1.

Take the first derivative of $f\left(x\right)=x\ast e^{x^2+1}$ using the Product and Chain rule.

$f=x$ and $f\prime=1$

$g=e^{x^2+1}$ and $g^\prime=e^{x^2+1}\ast2x$

Plug values into the formula:

$f^\prime\left(x\right)=f^\prime g+fg^\prime=$

$1\ast e^{x^2+1}+x\ast e^{x^2+1}\ast2x$

Step 2.

Plug in $x=2$ to find $f^\prime\left(2\right)$

$f^\prime\left(2\right)=1\ast e^{2^2+1}+2\ast e^{2^2+1}\ast2\ast2$

$f^\prime\left(2\right)=e^5+8e^5=9e^5$

The correct answer is $A$ .

Baruch MTH 2207 Final Topic 5. Chain Rule.

Chain Rule.

$f\left(x\right)=u^n\rightarrow f^\prime\left(x\right)=n\ast u^{n-1}\ast u\prime$

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 $(the u\prime)$ .

MTH 2207, Sample Final SP17, Problem 21

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEM 5

Step 1.

Make the U-substitution for $f\left(x\right)=\left(4x^2+1\right)^3$

$u=4x^2+1\rightarrow u^\prime=8x$

$f\left(x\right)=u^3\rightarrow f^\prime\left(x\right)=3u^2\ast u^\prime=3\left(4x^2+1\right)\ast8x=24x(4x^2+1)$

The answer is $E$ .

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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 $f\prime(x)$

Step 2.

Find $f\prime(x_0)$ – this is the slope

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 2207, Sample Final SP17, Problem 13

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEM 6
Let’s linearize the following function $y=2x^3+1$ at the point $(2,17)$

Step 1.

$f\left(x\right)=2x^3+1\rightarrow f^\prime\left(x\right)=6x^2$

Step 2.

$f^\prime(2)=6\ast2^2=24$

Step 3.

$f\left(2\right)=2\ast2^3+1=16+1=17$

Step 4.

$y=24\ast\left(x-2\right)+17$

$y=24x-48+17$

$y=24x-31$

The answer is $D$ .

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 $f\prime\prime(x)$

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 2207, Sample Final FA17, Problem 9

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEM 7
Let’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 $f^{\prime\prime}\left(x\right)=(3-x)(x^2-4)$ and we are asked to find the intervals on which this function is concave up.

Step 1.

$f^{\prime\prime}\left(x\right)=(3-x)(x^2-4)$

Let’s factor this function further: $f^{\prime\prime}\left(x\right)=(3-x)(x-2)(x+2)$

Step 2.

$f^{\prime\prime}\left(x\right)=0$

$\left(3-x\right)\left(x-2\right)\left(x+2\right)=0\rightarrow x=-2,x=2,\ x=3$

Step 3.

Let’s now test these inflection points by plugging values around them:

$f^{\prime\prime}\left(-3\right)=\left(3-\left(-3\right)\right)\left(\left(-3\right)^2-4\right)=6\ast\left(9-4\right)=30\rightarrow(+)$

We can conclude the following:

$\left(-\infty,-2\right)\rightarrow(+)$

$\left(-2,2\right)\rightarrow(-)$

$\left(2,3\right)\rightarrow(+)$

$\left(3,\infty\right)\rightarrow(-)$

The function is concave up for the following intervals: $\left(-\infty,-2\right) and \left(2,3\right)$ .

The answer is $A$ .

Baruch MTH 2207 Final Topic 8. Differentials.

Differential problems include proper computation of $dy$ and/or $dx$ .

Rule 1.

$f^\prime\left(x\right)=\frac{dy}{dx}\rightarrow dy=f^\prime\left(x\right)dx$

Rule 2.

$dy=f^\prime\left(x_{initial}\right)\ast(x_{final}-x_{initial})$

MTH 2207, Sample Final SP17, Problem 1

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEM 8
Differential problems can be simple if you know the rules. Let’s find the differential $dy$ for $y=x^\frac{3}{2}$ .

Step 1.

Take the first derivative of $y=x^\frac{3}{2}$

$\frac{dy}{dx}=\frac{3}{2}x^\frac{1}{2}$

$dy=\frac{3}{2}x^\frac{1}{2}\ast dx$

Step 2.

The differential of $y$ or $dy$ needs to be found.

Since $x$ is changing from $64$ to $64.1$ , the $dx=64.1-64=0.1$ and $x=64$ .

Let’s plug in these values:

$dy=\frac{3}{2}{64}^\frac{1}{2}\ast0.1=\frac{3}{2}\ast8\ast0.1=1.2$

The answer is $B$ .

Baruch MTH 2207 Final Topic 9. Implicit Differentiation.

In Implicit Differentiation problems, it is imperative to find $\frac{dy}{dx}$ . There are two simple rules for implicit differentiation problems.

Rule 1.

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

Rule 1.

Every time you take the derivative of $y$ , multiply the final result by $\frac{dy}{dx}$ .

MTH 2207, Sample Final SP17, Problem 22

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEM 9
Let’s find the slope of the tangent line to the following expression $3x^2+2xy-4y^2=12$ at the point $(2,1)$ .

Step 1.

Take the derivative of both sides of the expression:

( $3x^2+2xy-4y^2)\prime=(12)\prime$

a. $\left(3x\right)^\prime=6x$

b. $\left(2xy\right)^\prime:$

$f=2x\rightarrow f^\prime=2$

$g=y\rightarrow g^\prime=\frac{dy}{dx}$

Let’s plug in the values into the following expression:

$f^\prime g+fg^\prime=2y+2x\frac{dy}{dx}$

c. $\left(-4y^2\right)=-8y\frac{dy}{dx}$

$\left(3x^2+2xy-4y^2\right)^\prime=\left(12\right)^\prime$

$6x+2y+2x\frac{dy}{dx}-8y\frac{dy}{dx}=0$

Step 2.

Let’s now plug in the value $(2,1)$ into the first derivative expression:

$6\ast2+2\ast1+2\ast2\frac{dy}{dx}-8\ast1\frac{dy}{dx}=0$

$12+2+4\frac{dy}{dx}-8\frac{dy}{dx}=0$

$14-4\frac{dy}{dx}=0$

$4\frac{dy}{dx}=14$

$\frac{dy}{dx}=\frac{14}{4}\rightarrow\frac{dy}{dx}=\frac{7}{2}$

The answer is $E$ .

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 that’s the case, make the U-substitution for the bottom.

MTH 2207, Sample Final SP17, Problem 6

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEM 10

Step 1.

The more complex-looking expression in the fraction is $x^3+2x^2+x$

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

$u\prime=3x^2+4x+1$

Keep in mind that $u\prime=\frac{du}{dx}$ , so we can say the following:

$\frac{du}{dx}=3x^2+4x+1$

Solving for $dx$ we get the following:

$\frac{du}{3x^2+4x+1}=dx$

Step 2.

Plug in u and dx into the equation $\int{\frac{6x^2+8x+2}{x^3+2x^2+x}dx}$

$\int{\frac{6x^2+8x+2}{x^3+2x^2+x}dx}=$

$\int{\frac{6x^2+8x+2}{u}\ast\frac{du}{3x^2+4x+1}}=$

$\int{\frac{2}{u}\ast d u}=2\ln{\left|u\right|}+C$

Step 3.

We have to plug in the value for $u$ back $u=x^3+2x^2+x$

$2\ln{\left|x^3+2x^2+x\right|}+C$

The answer is $E$ .

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Baruch MTH 2207 Final Overview

Baruch MTH 2207 Final Topic 1. Limits and Continuity.

MTH 2207, Sample Final SP17, Problem 25

Step 1.

Baruch MTH 2207 Final Topic 2. Asymptotes.

Step 1.

Step 2.

Step 1.

Step 2.

Step 3.

MTH 2207, Sample Final SP17, Problem 5

Step 1.

Step 2.

Baruch MTH 2207 Final Topic 3. Derivative of Logarithms.

Rule 1.

Rule 2.

MTH 2207, Sample Final SP17, Problem 4

Step 1.

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

Product Rule.

Quotient Rule.

MTH 2207, Sample Final FA17, Problem 6

Step 1.

Step 2.

Baruch MTH 2207 Final Topic 5. Chain Rule.

Chain Rule.

MTH 2207, Sample Final SP17, Problem 21

Step 1.

Baruch MTH 2207 Final Topic 6. Slope and Linearization.

Step 1.

Step 2.

Step 3.

Step 4.

MTH 2207, Sample Final SP17, Problem 13

Step 1.

Step 2.

Step 3.

Step 4.

Baruch MTH 2207 Final Topic 7. Concavity.

Step 1.

Step 2.

Step 3.

MTH 2207, Sample Final FA17, Problem 9

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Step 2.

Step 3.

Baruch MTH 2207 Final Topic 8. Differentials.

Rule 1.

Rule 2.

MTH 2207, Sample Final SP17, Problem 1

Step 1.

Step 2.

Baruch MTH 2207 Final Topic 9. Implicit Differentiation.

Rule 1.

Rule 1.

MTH 2207, Sample Final SP17, Problem 22

Step 1.

Step 2.

Baruch MTH 2207 Final Topic 10. Antiderivatives.

MTH 2207, Sample Final SP17, Problem 6

Step 1.

Step 2.

Step 3.

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