Pre-Calculus - Math

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Question

The speed of a car traveling on the highway is given by the following function of time:

What can you say about the car's speed after a long time (that is, as approaches infinity)?

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Answer

The function given is a polynomial with a term , such that is greater than 1.

Whenever this is the case, we can say that the whole function diverges (approaches infinity) in the limit as approaches infinity.

This tells us that the given function is not a very realistic description of a car's speed for large !

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Question

$\textup{Find: }\lim_{x \to \infty }\frac{2x^{2}-7x}{x^{2}+3}$

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Answer

$\textup{As } x \to\infty,\textup{all but the terms with the highest exponents become insignificant.}$

$\textup{As }x^{2} \textup{ is the highest on top and bottom, evaluate the coefficients:}} \frac{2}{1}=2$

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Question

Calculate $\lim_{x\rightarrow \infty }\left ( x^{2} \sin \frac{1}{ x^{2}-1} - \sin \frac{1}{ x^{2}-1} \right )$ .

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Answer

This can be rewritten as follows:

$\lim_{x\rightarrow \infty }\left ( x^{2} \sin \frac{1}{ x^{2}-1} - \sin \frac{1}{ x^{2}-1} \right )$

$= \lim_{x\rightarrow \infty }\left ( x^{2}\cdot \sin \frac{1}{ x^{2}-1} -1\cdot \sin \frac{1}{ x^{2}-1} \right )$

$= \lim_{x\rightarrow \infty }\left [\left ( x^{2}-1 \right ) \cdot \sin \frac{1}{ x^{2}-1} \right]$

$= \lim_{x\rightarrow \infty } \frac{ \sin \frac{1}{ x^{2}-1} }{\frac{1}{ x^{2}-1}}$

We can substitute $u = \frac{1}{ x^{2}-1}$ , noting that as $x\rightarrow \infty$ , $u \rightarrow 0$ :

$= \lim_{u\rightarrow 0} \frac{ \sin u }{u} = 1$ , which is the correct choice.

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Question

Let $f(x)=\frac{1}{x}$ .

Find $\lim_{x\ \to \0^{-}} \frac{1}{x}$ .

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Answer

This is a graph of $\frac{1}{x}$ . We know that $\frac{1}{0}$ is undefined; therefore, there is no value for . But as we take a look at the graph, we can see that as approaches 0 from the left, approaches negative infinity.

This can be illustrated by thinking of small negative numbers.

NOTE: Pay attention to one-sided limit specifications, as it is easy to pick the wrong answer choice if you're not careful.

$\lim_{x \to \0^+} \frac{1}{x}$ is actually infinity, not negative infinity.

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Question

$f(x)=\frac{x^{2}-x-6}{x+2}$

$\lim_{x\rightarrow -2}f(x) =$

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Answer

A limit describes what -value a function approaches as approaches a certain value (in this case, ). The easiest way to find what -value a function approaches is to substitute the -value into the equation.

$f(-2)=\frac{(-2)^{2}-(-2)-6}{-2+2}=\frac{4+2-6}{0}=\frac{0}{0}$

Substituting for gives us an undefined value (which is NOT the same thing as 0). This means the function is not defined at that point. However, just because a function is undefined at a point doesn't mean it doesn't have a limit. The limit is simply whichever value the function is getting close to.

One method of finding the limit is to try and simplify the equation as much as possible:

$f(x)=\frac{(x-3)(x+2)}{x+2}$

As you can see, there are common factors between the numerator and the denominator that can be canceled out. (Remember, when you cancel out a factor from a rational equation, it means that the function has a hole -- an undefined point -- where that factor equals zero.)

After canceling out the common factors, we're left with:

Even though the domain of the original function is restricted ( cannot equal ), we can still substitute into this simplified equation to find the limit at

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Question

A function is defined by the following rational equation:

$f(x)=\frac{x+2}{2x+5}$

What are the horizontal and vertical asymptotes of this function's graph?

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Answer

To find the horizontal asymptote, compare the degrees of the top and bottom polynomials. In this case, the two degrees are the same (1), which means that the equation of the horizontal asymptote is equal to the ratio of the leading coefficients (top : bottom). Since the numerator's leading coefficient is 1, and the denominator's leading coefficient is 2, the equation of the horizontal asymptote is $y=\frac{1}{2}$ .

To find the vertical asymptote, set the denominator equal to zero to find when the entire function is undefined:

$x=-\frac{5}{2}$

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Question

Indicate the first three terms of the following series:

$\sum_{x=1}^{7}(4x-5)$

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Answer

In the arithmetic series, the first terms can be found by plugging , , and into the equation.

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Question

Indicate the first three terms of the following series:

$\sum_{c=3}^{10}(8-5c)$

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Answer

In the arithmetic series, the first terms can be found by plugging in , , and for .

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Question

Find the sum of all even integers from to .

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Answer

The formula for the sum of an arithmetic series is

$S = \frac{n}{2}(t_{1}+t_{n})$ ,

where is the number of terms in the series, $t_{1}$ is the first term, and $t_{n}$ is the last term.

$S=\frac{30}{2}(2+60)=15(62)=930$

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Question

Find the sum of all even integers from to .

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Answer

The formula for the sum of an arithmetic series is

$S = \frac{n}{2}(t_{1}+t_{n})$ ,

where is the number of terms in the series, $t_{1}$ is the first term, and $t_{n}$ is the last term.

We know that there are terms in the series. The first term is and the last term is . Our formula becomes:

$S = \frac{50}{2}(250+350)$

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Question

Find the sum of the even integers from to .

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Answer

The sum of even integers represents an arithmetic series.

The formula for the partial sum of an arithmetic series is

$S = \frac{n}{2}(2(a_1)+(n-1)d)$ ,

where is the first value in the series, is the number of terms, and is the difference between sequential terms.

Plugging in our values, we get:

$S = \frac{51}{2}(2(250)+(51-1)2)$

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Question

Find the value for $\sum_{i=0}^{\infty} (\frac{1}{2})^i$

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Answer

To best understand, let's write out the series. So

$\sum_{i=0}^{\infty} (\frac{1}{2})^i = 1+1/2+(1/2)^2 + (1/2)^3+...+(1/2)^\infty$

We can see this is an infinite geometric series with each successive term being multiplied by $\frac{1}{2}$ .

A definition you may wish to remember is

$1+r+r^2+r^3+... +r^\infty= 1/(1-r)$ where stands for the common ratio between the numbers, which in this case is $\frac{1}{2}$ or . So we get

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Question

Evaluate:

$1 - \frac{1}{6} + \frac{1}{36} - \frac{1}{216} +...$

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Answer

This is a geometric series whose first term is $a_{0} = 1$ and whose common ratio is $r = - \frac{1}{6}$ . The sum of this series is:

$\frac{a_{0}}{1-r} = \frac{1}{1- \left (- \frac{1}{6} \right )} = \frac{1}{\frac{6}{6}+ \frac{1}{6}}= \frac{1}{\frac{7}{6}} =\frac{6}{7}$

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Question

What is the domain of the function below?

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

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Answer

The domain is defined as the set of all values of x for which the function is defined i.e. has a real result. The square root of a negative number isn't defined, so we should find the intervals where that occurs:

$x^2-x+\frac{1}{4} <0 \Rightarrow (x-\frac{1}{2})^2<0$

The square of any number is positive, so we can't eliminate any x-values yet.

If the denominator is zero, the expression will also be undefined.

Find the x-values which would make the denominator 0:

$(x-\frac{1}{2})^2=0\Rightarrow (x-\frac{1}{2})=0\Rightarrow x=\frac{1}{2}$

Therefore, the domain is $x\neq \frac{1}{2}$ .

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Question

Evaluate:

$1 + \frac{1}{7} + \frac{1}{49} + \frac{1}{343} +...$

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Answer

This is a geometric series whose first term is $a_{0} = 1$ and whose common ratio is $r = \frac{1}{7}$ . The sum of this series is:

$\frac{a_{0}}{1-r} = \frac{1}{1- \frac{1}{7} } = \frac{1}{\frac{7}{7}- \frac{1}{7}}= \frac{1}{\frac{6}{7}} =\frac{7}{6}$

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Question

Consider the sequence: $2,\ 5,\ 8,\ 11,\ ...$

What is the fifteenth term in the sequence?

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Answer

The sequence can be described by the equation $\small 2+3(n-1)$ , where is the term in the sequence.

For the 15th term, .

$\small 2+3(n-1)=2+3(15-1)$

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Question

What is the sixth term when $\left (\frac{1}{2}x-2 \right )^{10}$ is expanded?

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Answer

We will need to use the Binomial Theorem in order to solve this problem. Consider the expansion of , where n is an integer. The rth term of this expansion is given by the following formula:

$\binom{n}{r-1}a^{n-r+1}b^{r-1}$ ,

where $\binom{n}{r-1}$ is a combination. In general, if x and y are nonnegative integers such that x > y, then the combination of x and y is defined as follows: $\binom{x}{y}=_xC_y=\frac{x!}{(y)!(x-y)!}$ .

We are asked to find the sixth term of $\left (\frac{1}{2}x-2 \right )^{10}$ , which means that in this case r = 6 and n = 10. Also, we will let $a=\frac{1}{2}x$ and . We can now apply the Binomial Theorem to determine the sixth term, which is as follows:

$\binom{10}{6-1}\left (\frac{1}{2}x \right )^{10-6+1}\cdot(-2) ^{6-1}$

$=\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$

Next, let's find the value of $\binom{10}{5}$ . According to the definition of a combination,

$\binom{10}{5}=\frac{10!}{5!5!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5!}{5!5!}$

$=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2}=252$ .

Remember that, if n is a positive integer, then $n!=n\cdot(n-1)\cdot(n-2)\cdot\cdot\cdot3\cdot2\cdot1$ . This is called a factorial.

Let's go back to simplifying $\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$ .

$\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}=252\cdot\left (\frac{1}{2}x \right )^{5}\cdot-32$

$=252\cdot\left (\frac{1}{2} \right )^5\cdot x^5\cdot (-32) =252\cdot \frac{1}{32}\cdot-3 2\cdot x^5$

The answer is .

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Question

What are the first three terms in the series?

$\sum_{n=1}^{7}(4n-5)$

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Answer

To find the first three terms, replace with , , and .

The first three terms are , , and .

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Question

Find the first three terms in the series.

$\sum_{n=3}^{10}(8-5n)$

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Answer

To find the first three terms, replace with , , and .

The first three terms are , , and .

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Question

Indicate the first three terms of the following series:

$\sum_{m=-2}^{9}(2)^m$

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Answer

The first terms can be found by substituting , , and for into the sum formula.

$(2)^{-2}=\frac{1}{4}$

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

$(2)^{0}=1$

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