Crush Baruch College MTH 2205 - Final Test SP17 In Under 40 Minutes Or LESS

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

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:

$dy=y^\prime\ast dx$

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

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 $\bigcup f o r m$ 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.

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

The boundaries are $x=1$ and $x=4$

There are $3$ rectangles, so the width of each rectangle is:

$rectangle\ width=\frac{4-1}{3}=1$

Since we are looking at the RIGHT-hand approximation, the points are $2$ , $3$ , and $4$ .

Step 2.

Plug in the values into the following equations

$1\ast f\left(2\right)+1\ast f\left(3\right)+1\ast f\left(4\right)$

Let’s find the values of individual elements:

$f\left(2\right)=2^2+2=6$

$f\left(3\right)=3^2+2=11$

$f\left(4\right)=4^2+2=18$

$1\ast6+1\ast11+1\ast18=6+11+18=35$

The correct 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

The formula for the elasticity of demand is $E=\frac{p}{x}\ast\frac{dx}{dp}$

Step 1.

Based on the formula above, we need to have values for $p$ , $x$ , and $\frac{dx}{dp}$ .

We already know that $p=10$ . Let’s find the value for $x=10$ :

$x=1000-4\ast{10}^2=1000-400=600$

Step 2.

Let’s find $\frac{dx}{dp}$

$\frac{dx}{dp}\ =-8p$

Step 3.

Let’s compute $\frac{dx}{dp}$ , when $p=10$ :

$\frac{dx}{dp}\ =-8\ast10=-80$

Step 4.

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

$E=\frac{10}{600}\ast-80=-\frac{800}{600}=-\frac{4}{3}$

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:

$Average\ Value=\frac{1}{b-a}\ast\int_{1}^{4}\frac{1}{\sqrt x}dx$

Step 1.

Let’s first rewrite the equation under the integral and find the integral:

$\int\frac{1}{\sqrt x}dx=\int\frac{1}{x^\frac{1}{2}}dx=\int x^{-\frac{1}{2}}dx=\frac{x^\frac{1}{2}}{\frac{1}{2}}=2\sqrt x$

Step 2.

Simply plug the values into the Average Value function:

$Average\ Value=\frac{1}{4-1}\ast\int_{1}^{4}\frac{1}{\sqrt x}dx=$

$\frac{1}{4-1}\ast2\sqrt x|_1^4=\frac{1}{3}\ast\left(2\sqrt4-2\sqrt1\right)=\frac{1}{3}\ast\left(4-2\right)=\frac{2}{3}$

The answer is $E$ .

Problem 9

Step 1.

Take the first derivative of $y=3x^2+kx-5$

$y^\prime=6x+k$

Step 2.

Let’s set up the first derivative to $0$ , plug in $x=-2$ and solve for $k$ :

$0=6\ast(-2)+k$

$0=-12+k\rightarrow k=12$

The answer is $C$ .

Problem 10

Keep in mind that $s(t)$ is the anti-derivative of $v(t)$

Step 1.

Take the integral of $v\left(t\right)=3t^2+t$ :

$s\left(t\right)=\int{(3t^2}+t)\ dt=t^3+\frac{t^2}{2}+C$

Step 2.

Plug in $t=2$ and solve for $C$ :

$s\left(2\right)=2^3+\frac{2^2}{2}+C=3$

$8+\frac{4}{2}+C=3$

$8+2+C=3\rightarrow C=-7$

Step 3.

Plug in $C=-7$ for the integral computed above:

$s\left(t\right)=t^3+\frac{t^2}{2}-7$ , and the answer is $A$ .

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Crush Baruch College MTH 2205 – Final Test SP17 In Under 40 Minutes Or LESS

Baruch College MTH 2205 – Final Test SP17 Overview

Problem 1

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Problem 2

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Problem 3

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Problem 7

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Problem 8

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Problem 9

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Problem 10

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Baruch College MTH 2205 – Final Test SP17 Overview

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Problem 9

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Problem 10

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