Functions, Graphs, and Limits - AP Calculus AB

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Question

$f(x)={x^{2}+3}$ when and

$f(x)={7x-3}$ when $x\geq 3$

At the funciton described above is:

Answer

The answer is both.

If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.

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Question

Which of the following functions contains a removeable discontinuity?

Answer

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.

For example, the function $f(x)=\frac{1+x^3}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:

$f(x)=\frac{1+x^3}{1+x}=\frac{(1+x)(x^2-x+1)}{1+x}=x^{2}-x+1$ , where $x eq -1$ .

Thus, we could say that $\lim_{x\rightarrow -1}\frac{1+x^3}{1+x}=\lim_{x\rightarrow -1}x^2-x+1=(-1)^2-(-1)+1=3$ .

As we can see, the limit of exists at , even though is undefined.

What this means is that will look just like the parabola with the equation $x^{2}-x+1$ EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

$f(x)=\frac{x}{1-x^{2}}$ , and

$f(x)=1-(\ln x)^{2}$

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

$f(x)=x^{2}\sin x^{2}$ and $f(x)=\frac{x+1}{1+x^{2}}$ are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

$f(x)=\frac{1+x^{3}}{1+x}$ .

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Question

Is the following piecewise function continuous for all ? If not, state where it is discontinuous.

$f(x) = \left{\begin{Bmatrix} x^2+3 & 3 < x \ 5x-3 & 3 \leq x < 5 \ 11(x-3) &x \geq 5 \ \end{matrix}\right.}$

Answer

To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.

At , this means checking that and have the same value. More formally, we are checking to see that $f(3) = \lim_{x\rightarrow3^-} f(x)$ , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.

Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.

Next, for , we have and . Plugging in 5, we get 22 for both equations.

As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.

Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.

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Question

$f(x)={x^{2}+3}$ when and

$f(x)={7x-3}$ when $x\geq 3$

At the funciton described above is:

Answer

The answer is both.

If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.

Compare your answer with the correct one above

Question

Which of the following functions contains a removeable discontinuity?

Answer

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.

For example, the function $f(x)=\frac{1+x^3}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:

$f(x)=\frac{1+x^3}{1+x}=\frac{(1+x)(x^2-x+1)}{1+x}=x^{2}-x+1$ , where $x eq -1$ .

Thus, we could say that $\lim_{x\rightarrow -1}\frac{1+x^3}{1+x}=\lim_{x\rightarrow -1}x^2-x+1=(-1)^2-(-1)+1=3$ .

As we can see, the limit of exists at , even though is undefined.

What this means is that will look just like the parabola with the equation $x^{2}-x+1$ EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

$f(x)=\frac{x}{1-x^{2}}$ , and

$f(x)=1-(\ln x)^{2}$

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

$f(x)=x^{2}\sin x^{2}$ and $f(x)=\frac{x+1}{1+x^{2}}$ are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

$f(x)=\frac{1+x^{3}}{1+x}$ .

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Question

Is the following piecewise function continuous for all ? If not, state where it is discontinuous.

$f(x) = \left{\begin{Bmatrix} x^2+3 & 3 < x \ 5x-3 & 3 \leq x < 5 \ 11(x-3) &x \geq 5 \ \end{matrix}\right.}$

Answer

To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.

At , this means checking that and have the same value. More formally, we are checking to see that $f(3) = \lim_{x\rightarrow3^-} f(x)$ , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.

Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.

Next, for , we have and . Plugging in 5, we get 22 for both equations.

As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.

Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.

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Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align}$

Answer

$\begin{align}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-1.5\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(18x + 17)}{(12x - 11)}\end{align}$

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Question

Find all vertical asymptotes and horizontal asymptotes of the function,

$f(x) = \frac{1-2x^3}{x^3-4x^2+4x}$

Answer

$f(x) = \frac{1-2x^3}{x^3-4x+2x}$

1) To find the horizontal asymptotes, find the limit of the function as $x->\infty$ ,

$\lim_{x->\infty}\frac{1-2x^3}{x^3-4x^2+4x}=\lim_{x->\infty}\frac{\frac{1}{x^3}-2}{1-\frac{4}{x}+\frac{4}{x^2}} = -2$

Therefore, the function has a horizontal asymptote

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2) Vertical asympototes will occur at points where the function blows up, $y->\infty$ . For rational functions this behavior occurs when the denominator approaches zero.

Factor the denominator and set to zero,

$x = \left { 0, 2\right }$

So the graph of has two vertical asymptotes, one at and the other at . They have been drawn into the graph of below. The blue curves represent .

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Question

Evaluate

$\lim_{x\rightarrow \infty}\frac{4x^3+2x^2-5x+1}{x^3-x^2+3x-5}$

Answer

The equation $\frac{4x^3+2x^2-5x+1}{x^3-x^2+3x-5}$ will have a horizontal asymptote y=4.

We can find the horizontal asymptote by looking at the terms with the highest power.

The terms with the highest power here are in the numerator and in the denominator. These terms will "take over" the function as x approaches infinity. That means the limit will reach the ratio of the two terms.

The ratio is $\frac{4x^3}{x^3}=\frac{4}{1}=4$

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Question

$\begin{align}&\text{Of the following functions, which most likely represents the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-1.8\&\text{We find a zero denominator for the function: }\&-\frac{(10x + 11)}{(11x + 20)}\end{align}$

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Question

$\lim_{x\rightarrow -\infty}\frac{4x^5-3x^2+2}{2x^3-x^2+1}=?$

Answer

For this infinity limit, we need to consider the leading terms of both the numerator and the denominator. In our problem, the leading term of the numerator is larger than the leading term of the denominator. Therefore, it will be growing at a faster rate.

$\lim_{x\rightarrow -\infty}\frac{4x^5-3x^2+2}{2x^3-x^2+1}=\frac{4x^5}{2x^3}=2x^2.$

Now, simply input the limit value, and interpret the results.

$\lim_{x\rightarrow -\infty} 2x^2=2(-\infty)^2=2*\infty=\infty.$

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Question

$\lim_{x\rightarrow \infty}\frac{2x^2+3x}{10x^2+x}$

Answer

For infinity limits, we need only consider the leading term in both the numerator and the denominator. Here, we have the case that the exponents are equal in the leading terms. Therefore, the limit at infinity is simply the ratio of the coefficients of the leading terms.

$\lim_{x\rightarrow \infty}\frac{2x^2+3x}{10x^2+x}=\frac{2x^2}{10x^2}=\frac{2}{10}=\frac{1}{5}.$

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Question

$\lim_{x\rightarrow \infty}\frac{3x^7+4x^4-3x^2}{5x^5+2x^4}$

Answer

Infinity limits can be found by only considering the leading term in both the numerator and the denominator. In this problem, the numerator has a higher exponent than the denominator. Therefore, it will keep increasing and increasing at a much faster rate. These limits always tend to infinity.

$\lim_{x\rightarrow \infty}\frac{3x^7+4x^4-3x^2}{5x^5+2x^4}=\frac{3x^7}{5x^5}=\frac{3}{5}x^2.$

$\lim_{x\rightarrow \infty}\frac{3}{5}x^2=\infty.$

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Question

$\lim_{x\rightarrow -\infty} \frac{2x^3-3x^2+4x}{4x^5-3x^3}$

Answer

For infinity limits, we only consider the leading term in both the numerator and the denominator. Then, we need to consider the exponents of the leading terms. Here, the denominator has a higher degree than the numerator. Therefore, we have a bottom heavy fraction. Even though we are evaluating the limit at negative infinity, this will still tend to zero because the denominator is growing at a faster rate. You can convince yourself of this by plugging in larger and larger negative values. You will just get a longer and smaller decimal.

$\lim_{x\rightarrow -\infty} \frac{2x^3-3x^2+4x}{4x^5-3x^3}=\frac{2x^3}{4x^5}=\frac{1}{2x^2}$

$\lim_{x\rightarrow -\infty}\frac{1}{2x^2}=0.$

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Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align}$

Answer

$\begin{align}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=0.75\&\text{This matches the ratio of coefficients for the function: }\&\frac{(6x + 1)}{(8x - 17)}\end{align}$

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Question

$\begin{align}&\text{Which of the four following functions is depicted in the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-1.1\&\text{We find a zero denominator for the function: }\&\frac{(8x - 2)}{(10x + 11)}\end{align}$

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Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align}$

Answer

$\begin{align}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-0.3\&\text{We find a zero denominator for the function: }\&-\frac{(x - 11)}{(4x + 1)}\end{align}$

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Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align}$

Answer

$\begin{align}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-2.67\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(8x - 18)}{(3x + 3)}\end{align}$

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Question

$\begin{align}&\text{Which of the four following functions is depicted in the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-0.8\&\text{We find a zero denominator for the function: }\&\frac{(10x - 1)}{(6x + 5)}\end{align}$

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Question

$\begin{align}&\text{Of the following functions, which most likely represents the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=2\&\text{This matches the ratio of coefficients for the function: }\&\frac{(16x - 12)}{(8x - 18)}\end{align}$

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