TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEMS

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Baruch College MTH 2003 Final Overview

MTH 2003 is a precalculus course in Baruch College. MTH 2003 is a preliminary mathematics course students must take prior to starting the calculus as well as in quantitative courses in allied disciplines track. This course is a prerequisite for MTH 2205. Baruch College MTH 2003 Final Test has a calculator section. 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 2003 CLICK HERE

To get your hands on Baruch College MTH 2003 Last Minute Cram Course, CLICK HERE

To view sample questions for Baruch MTH 2003 Final CLICK HERE.

Baruch MTH 2003 Final Topic 1. The Line.

Rule 1.

Slopes of parallel lines are equivalent.

Rule 2.

Slopes of perpendicular lines are negative reciprocals of each other.

MTH 2003, Sample Final SP17, Problem 1

TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEM 1
We need to find the slope of a perpendicular line to the following linear function $3x-4y+1=0$ .

Step 1.

Solve for $y$ :

$3x-4y+1=0$

$4y=3x+1$

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

Step 2.

Slopes of perpendicular lines are negative reciprocals of each other.

The slope of the existing function is $\frac{3}{4}$ .

The negative reciprocal of the line is $-\frac{4}{3}$ .

The answer is $A$ .

Baruch MTH 2003 Final Topic 2. Parabola.

Quadratic equations and parabolas are the same thing. A quadratic equation looks like a parabola and every parabola can be expressed as a quadratic equation.

The final test often includes problems on vertex for functions $f\left(x\right)=ax^2+bx+c$

Step 1.

$x_{vertex}=-\frac{b}{2a}$

Step 2.

$y_{vertex}=f(x_{vertex})$

MTH 2003, Sample Final SP17, Problem 3

TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEM 2
Let’s solve a specific problem on vertex. In Problem 3 of the Sample Final SP17, we need to find the vertex for $f\left(x\right)=x^2-10x+13$

Step 1.

$x_{vertex}=-\frac{-10}{2\ast1}=5$

Step 1.

$y_{vertex}=5^2-10\ast5+13=25-50+13=-12$

The coordinate pair of the vertex is $(5,-12)$ .

The answer is $E$ .

Baruch MTH 2003 Final Topic 3. Circle.

Equation of a circle with radius $R$ and origin $(h,k)$ is $\left(x-h\right)^2+\left(y-k\right)^2=R^2$ . Complete the square for $x$ and $y$ to turn the equation into the form $\left(x-h\right)^2+\left(y-k\right)^2=R^2$ .

MTH 2003, Sample Final SP17, Problem 4

TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEM 3
In Problem 4 of the Final Test SP17 we need to find the radius of the circle for $x^2+y^2+14x-2y=0$

Step 1.

Rearrange the equation to move $x$ and $y$ terms together.

$x^2+y^2+14x-2y=0$

$x^2+14x+y^2-2y=0$

$x^2+14x+__+y^2-2y+__=0$

Step 2.

Let’s now complete the squares for $x$ and $y$ :

$x^2+14x+\left(\frac{14}{2}\right)^2+y^2-2y+\left(\frac{2}{2}\right)^2=\left(\frac{14}{2}\right)^2+\left(\frac{2}{2}\right)^2$

$x^2+14x+49+y^2-2y+1=49+1$

$\left(x+7\right)^2+\left(y-1\right)^2=50$

$R^2=50 and R=\sqrt{50}$

The answer is $C$ .

Baruch MTH 2003 Final Topic 4. Horizontal Asymptote.

Horizontal asymptote problems are typically very straight-forward. Identify the highest exponent with the coefficient on the top and on the bottom.

Step 1.

If the exponent on top is higher than the exponent on the bottom, there are no horizontal asymptotes.

Step 2.

If the exponent on top is equal to the exponent on the bottom, the horizontal asymptote is equal to the fraction of coefficients.

Step 3.

If the exponent on top is less than the exponent on the bottom, the horizontal asymptote is equal to $0$ .

Keep in mind that if the horizontal asymptote is, say, $10$ , the equation for the horizontal asymptote is $y=10$ .

MTH 2003, Sample Final SP17, Problem 5

TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEM 4
Let’s find the horizontal asymptote for $y=\frac{x-7}{x^2-49}$ .

Step 1.

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

The highest exponent of the numerator $x-7$ is $1$ .

The highest exponent of the denominator $x^2-49$ is $2$ .

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

The answer is $A$ .

Baruch MTH 2003 Final Topic 5. Limits.

Let’s consider limit problems of the following type: $\lim\below{x\rightarrow a}{f(x)}$ . Plug in $x=a$ into $f(x)$ right away:

Step 1.

If you get $\frac{0}{0}$ simplify top against the bottom and plug in $x=a$ again

Step 2.

If you get $\frac{0}{something}$ then the limit is equal to $0$

Step 3.

If you get $\frac{something}{0}$ then the limit is undefined

MTH 2003, Sample Final SP17, Problem 11

TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEM 5
We need to find the following limit $\lim\below{x\rightarrow3}{\frac{x-3}{x^2+4x-21}}$ .

Step 1.

Let’s plug in $x=3$ into the fraction right away: $\lim\below{x\rightarrow3}{\frac{3-3}{3^2+4\ast3-21}=\lim\below{x\rightarrow3}{\frac{0}{0}}}$

Step 2.

Since we got $\frac{0}{0}$ , we know that the fraction can be definitely simplified

$\lim\below{x\rightarrow3}{\frac{x-3}{x^2+4x-21}}=\lim\below{x\rightarrow3}{\frac{x-3}{(x-3)(x+7)}=\lim\below{x\rightarrow3}{\frac{1}{(x+7)}}}$

Let’s plug in $x=3$ again:

$\lim\below{x\rightarrow3}{\frac{1}{(x+7)}}=\frac{1}{3+7}=\frac{1}{10}$

The answer is $B$ .

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Baruch MTH 2003 Final Topic 6. 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 2003, Sample Final SP17, Problem 12

TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEM 6
In the Baruch MTH 2207 final SP17 test the following piece-wise function $f\left(x\right)=\left\{\begin{matrix}x^2-c,\ x\le5\\cx+6,\ x>5\\\end{matrix}\right.$ needs to become continuous on the entire interval $(-\infty,\infty)$ .

Step 1.

The border takes place at $x=5$ , so all we need to do is to plug it in into both equations and solve for $c$ :

$x^2-c=xc+6$

$5^2-c=5c+6$

$25-c=5c+6$

$19=6c$

$c=\frac{19}{6}$

Baruch MTH 2003 Final Topic 7. Economic Functions.

There are a couple of very useful formulas for economic function problems. If you know these formulas, you should be all set:

Revenue

$R\left(x\right)=px$ , where $p$ is demand and $x$ is the production level

Cost

$C\left(x\right)=Variable\ Cost\ast x+Fixed\ Cost$ , where $x$ is the production level

Profit

$P\left(x\right)=R\left(x\right)-C(x)$ , where $R(x)$ is revenue and $C(x)$ is the cost

MTH 2003, Sample Final SP17, Problem 6

TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEM 7
Great example to show the proper use of aforementioned formulas. The demand $p=100-2x$ and we need to find the number of items to be sold to maximize revenue.

Step 1.

The revenue function is

$R\left(x\right)=px$

$R\left(x\right)=\left(100-2x\right)x=100x-2x^2$

Step 2.

To find the number of items that will maximize revenue, take the first derivative

$R\prime(x)=100-4x$

Step 3.

Set the denominator to $0$ and solve for $x$

$R^\prime\left(x\right)=100-4x=0$

$x=25$

The answer is $D$ .

Baruch MTH 2003 Final Topic 8. Derivative.

Derivatives is a very extensive topic. But we will do the basic version of derivatives in this section of the article.

Basic Derivative
$f\left(x\right)=x^n\rightarrow f^\prime\left(x\right)=nx^{n-1}$

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

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

MTH 2003, Sample Final SP17, Problem 8

TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEM 8
We need to find the derivative of $f\left(x\right)=\frac{x+1}{x}$ . We are dealing with a fraction so, clearly, we are looking for the Quotient Rule.

Step 1.

Let’s assign the values for $f$ and $g$ , respectively:

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

$g=x$ and $g\prime=1$

Plug values into the formula:

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

$\frac{1\ast x-\left(x+1\right)\ast1}{x^2}=$

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

Step 2.

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

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

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

The answer is $D$ .

Baruch MTH 2003 Final Topic 9. 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 2003, Sample Final SP17, Problem 15

TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEM 9

Step 1.

Make the U-substitution for $f\left(x\right)=\sqrt{x^2+1}=\left(x^2+1\right)^\frac{1}{2}$

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

$f\left(x\right)=u^\frac{1}{2}\rightarrow$

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

Let’s make proper substitutions for $u$ and $u\prime$ :

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

The answer is $D$ .

Baruch MTH 2003 Final Topic 10. Horizontal Tangent Line.

If you do not know what to do on the Baruch MTH 2003 final, take the first derivative 😀. Horizontal tangent lines are not the exception to this rule of thumb.

In order to find the points at which the first derivative is equal to $0$ follow these simple steps:

Step 1.

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

Step 2.

Set the first derivative $f\prime\left(x\right)=0$

MTH 2003, Sample Final SP17, Problem 7

TOP 10 MUST KNOW BARUCH MTH 2003 FINAL PROBLEM 10
Let’s find the points at which horizontal tangent lines for $f\left(x\right)=-4x^2+32x-10$ are equal to $0$ .

Step 1.

$f^\prime(x)=-8x+32$

Step 2.

$f^\prime\left(x\right)=0\rightarrow-8x+32=0\rightarrow8x=32\rightarrow x=4$

Step 3.

Let’s now find the y value by plugging $x=4$ into the original function:

$f\left(4\right)=-4\ast4^2+32\ast4-10=-64+128-10=54$

The tangent line is horizontal at $(4,54)$ .

The answer is $A$ .

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Baruch College MTH 2003 Final Overview

Baruch MTH 2003 Final Topic 1. The Line.

Rule 1.

Rule 2.

MTH 2003, Sample Final SP17, Problem 1

Step 1.

Step 2.

Baruch MTH 2003 Final Topic 2. Parabola.

Step 1.

Step 2.

MTH 2003, Sample Final SP17, Problem 3

Step 1.

Step 1.

Baruch MTH 2003 Final Topic 3. Circle.

MTH 2003, Sample Final SP17, Problem 4

Step 1.

Step 2.

Baruch MTH 2003 Final Topic 4. Horizontal Asymptote.

Step 1.

Step 2.

Step 3.

MTH 2003, Sample Final SP17, Problem 5

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Baruch MTH 2003 Final Topic 5. Limits.

Step 1.

Step 2.

Step 3.

MTH 2003, Sample Final SP17, Problem 11

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Step 2.

Baruch MTH 2003 Final Topic 6. Continuity.

MTH 2003, Sample Final SP17, Problem 12

Step 1.

Baruch MTH 2003 Final Topic 7. Economic Functions.

Revenue

Cost

Profit

MTH 2003, Sample Final SP17, Problem 6

Step 1.

Step 2.

Step 3.

Baruch MTH 2003 Final Topic 8. Derivative.

MTH 2003, Sample Final SP17, Problem 8

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Step 2.

Baruch MTH 2003 Final Topic 9. Chain Rule.

MTH 2003, Sample Final SP17, Problem 15

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Baruch MTH 2003 Final Topic 10. Horizontal Tangent Line.

Step 1.

Step 2.

MTH 2003, Sample Final SP17, Problem 7

Step 1.

Step 2.

Step 3.

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