Defeat Baruch College MTH 2207 - Final Test SP17 In 40 Minutes Or Less

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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 $y=x^\frac{3}{2}$

$y^\prime=\frac{3}{2}x^\frac{1}{2}$

Step 2.

Remember that $y^\prime=\frac{dy}{dx}$

It means that

$dy=y^\prime\ast dx$

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

Step 3.

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 $B$ .

Problem 2

Remember that the function has critical numbers when its first derivative is equal to $0$ or does not exist.

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 $D$ .

Problem 3

Let’s analyze both conditions presented to us:

$f^\prime\left(x\right)>0$ and $f^{\prime\prime}\left(x\right)<0$ $f^\prime\left(x\right)>0$ means that the graph of the original function will be always increasing

$f^{\prime\prime}\left(x\right)<0$ means that the graph of the original function will be following the concave down concavity. Concave up concavity has a $\bigcup f o r m form$ and concave down concavity has the $\bigcap f o r m$ .

The only multiple choice that is both increasing and has a $\bigcap f o r m$ is option $D$ , which is the correct answer.

Problem 4

Keep in mind that the confusing $\frac{dy}{dx}$ term is just the first derivative.

Step 1.

Take the first derivative of $f\left(x\right)=xln\left(x^2\right)$ 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 $A$ .

Problem 5

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 $A$ .

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

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

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

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 $E$ .

Problem 7

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

Step 1.

Let’s take the derivative of both sides first:

$(V=\frac{4}{3}\pi r^3)\prime$

$\frac{dV}{dt}=\frac{4}{3}\pi\ast3r^2\frac{dr}{dt}\$

Step 2.

Plug in the values for $\frac{dV}{dt}=7$ and $r=10$ :

$7=\frac{4}{3}\pi\ast3{\ast10}^2\frac{dr}{dt}$

$7=\frac{4}{3}\pi\ast\frac{3}{1}\ast\frac{100}{1}\ast\frac{dr}{dt}$

$7=\frac{1200\pi}{3}\ast\frac{dr}{dt}$

$7=400\pi\ast\frac{dr}{dt}\rightarrow\frac{dr}{dt}=\frac{7}{400\pi}$

The answer is $B$ .

Problem 8

The average value of $f\left(x\right)=\frac{1}{\sqrt x}$ on the closed interval $\left[1,4\right]$ can be found by the same formula:
$Average\ Value=\frac{1}{b-a}\ast\int_{1}^{4}\frac{1}{\sqrt x}dx$