$\displaystyle f(x)$ is not differentiable at $\displaystyle x=0$ and $\displaystyle x=3$ because the values are discontinuities. $\displaystyle f(x)$ is not differentiable at $\displaystyle x=2$ because that point is a corner, indicating that the one-side limits at $\displaystyle f(2)$ are different. $\displaystyle f(1)$ is differentiable:the one side limits are the same and the point is continuous. 

Note that $\displaystyle f'(0) = 0$, indicating that there is a horizontal tangent on $\displaystyle f(x)$ at $\displaystyle x = 0$. More specifically, the derivative is the slope of the tangent line. If the slope of the tangent line is 0, then the tangent is horizontal.

The other two are incorrect because sharp turns only apply when we want to take the derivative of something. The derivative of a function at a sharp turn is undefined, meaning the graph of the derivative will be discontinuous at the sharp turn. (To see why, ask yourself if the slope at $\displaystyle x = 0$ is positive 1 or negative 1?) On the other hand, integration is less picky than differentiation: We do not need a smooth function to take an integral.

In this case, to get from $\displaystyle f'(x)$ to $\displaystyle f(x)$, we took an integral, so it didn't matter that there was a sharp turn at the specified point. Thus, neither function had any discontinuities. 

To solve this, find where the function cannot exist. Here, the function cannot exist if the denominator is zero. This happens at x=2 and x=-2. Graph the function on a graphing calculator or by hand to see that the function never crosses these vertical lines. It only gets infinitely close. This is characteristic of vertical asymptotes. 

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

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

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

 Therefore, the function $\displaystyle f(x)$ has a horizontal asymptote $\displaystyle y = -2$

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

Factor the denominator and set to zero, 

$\displaystyle x(x^2-4x+4)=x(x-2)^2 = 0$

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

So the graph of $\displaystyle f(x)$ has two vertical asymptotes, one at $\displaystyle x = 0$ and the other at $\displaystyle x = 2$.  They have been drawn into the graph of  $\displaystyle f(x)$ below. The blue curves represent $\displaystyle f(x)$. 

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. 

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

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

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

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. 

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

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.

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

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

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.

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

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

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

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