Special Functions

Special Functions exercise

• Decide which graph belongs to which function.

  Hints

  You can recognize the absolute value function by the v-shape.

  With the function $f(x)=\frac 1x$, for increasing values of $x$, you get decreasing values of $y$. Thus we have the $x$-axis as an asymptote.

  Solution

  You can recognize the absolute value function $f(x)=|x|$ by the v-shape. So that's the graph on the left above.

  The next graph belongs to $f(x)=\frac1x$. This graph has the $x$-axis as an asymptote. This means that as the $x$-value gets larger, the $y$-value gets closer and closer to zero.

  The graph on the left below belongs to $f(x)=\sqrt x$. Just put some values for $x$ in and check the resulting points; for instance, check $(1,1)$, $(4,2)$.

  The periodic graph on the right below is the graph of the tangent function: $f(x)=\tan(x)$.

• Write the definition of an asymptote.

  Hints

  Here you see an example: the function $f(x)=\frac1x$, with the $x$-axis as a horizontal asymptote and the $y$-axis as a vertical asymptote.

  The tangent function has vertical asymptotes too.

  Solution

  What is an asymptote?

  An asymptote is an invisible line which a curve, i.e. the graph of a function, comes closer and closer to but will never touch.

  For example, consider $f(x)=\frac1x$.

  Putting increasing values in $f(x)$ for $x$ gives:

  • $f(1)=1$
  • $f(2)=\frac12=0.5$
  • $f(3)=\frac13$
  • ...
  • $f(10)=\frac1{10}=0.1$
  • ...
  • $f(100)=\frac1{100}=0.01$

  We can then see that the function values approach zero, but that there does not exist any $x$ with $\frac1x=0$.

  Thus the $x$-axis is a horizontal asymptote of the corresponding graph.

• Determine the corresponding function equation.

  Hints

  If you see a v-shape, then it must be an absolute value function.

  You can also check asymptotes.

  Both v-shapes are only moved up or down.

  Pay attention to the reverse v-shape of the green graph.

  You can also check some points lying on the graph.

  Solution

  Both graphs on the right have a v-shape. So they belong to absolute value functions.

  The green one has the vertex at $y=4$. This graph is a reverse v-shape. Thus we have the function $f(x)=-|x|+4$.

  In a same manner we can handle the red graph: The vertex lies at $y=2$ and it's a direct v-shape. This gives us $f(x)=|x|+2$.

  Next we look at the graphs with asymptotes: the pink one and the purple one.

  The pink one is symmetric about the $y$-axis, with asymptotes at $y=0$ and $x=0$. So this must be the function $f(x)=\frac1{|x|}$.

  For the purple graph, we can see that the point $(1,2)$ lies on it and has asymptotes at $y=0$ and $x=0$. The function with these characteristics is $f(x)=\frac2x$.

• Explain the expressions' domain and range.

  Hints

  The domain is the set of all possible values for $x$. Check if there exists any values for $x$ which you can't put in the function equation.

  The range is the set of all possible function values.

  Let's have a look at the following example: $f(x)=x^2$.

  You can square any $x$-value, thus the domain is $D=\mathbb{R}$, all real numbers.

  But independent of which value for $x$ you put in $f(x)$, the function value is greater than or equal to zero. So the range is $R=\{y~|~y\ge 0\}$.

  Solution

  To find the domain, we have to see which values of $x$ can we put in $f(x)$ to get a well-defined output. For the range we have to determine all possible function values.

  $\begin{array}{l|c|c|c|c|c} F(x)=\rightarrow & |x| & ( \frac1x ) & \sqrt{x} & \sin{x} & \cos{x} \\ \hline \text{Domain:} &\mathbb{R} & x~|~x\in\mathbb{R};~x\neq 0 & x \ge 0 & \mathbb{R} & \mathbb{R} \\ \hline \text{Range:} &f(x) \ge 0 & \mathbb{R}, f(x) \neq 0 & f(x) \ge 0 & -1 \le f(x) \le 1 &-1 \le f(x) \le 1 \end{array}$

  $~$

  • Let's start with $f(x)=|x|$. Sure we can put each real number in for $x$. Thus $D=\{x~|~x\in\mathbb{R}\}$. The absolute value is defined as the distance of any number to zero and so it's positive for any $x\neq 0$. The absolute value of $|0|=0$. This gives us the range $R=\{y~|~y\ge 0\}$.

  • Next $f(x)=\frac1x$. Dividing by zero isn't possible at all. So we have to exclude $0$ from the domain: $D=\{x~|~x\in\mathbb{R};~x\neq 0\}$. The function $f(x)=\frac1x$ gives us any real number except $0$, because $\frac1x\neq0$ for all $x$. So we can conclude that also the range is $R=\{y~|~y\in\mathbb{R};~y\neq 0\}$.

  • The radical is the inversion of the square. The domain of $f(x)=\sqrt x$ is given by $D=\{x~|~x\ge 0\}$. The range is the set of all real number either positive or equal to zero: $R=\{y~|~y\ge 0\}$.

  • Last we investigate $f(x)=\sin(x)$ and $g(x)=\cos(x)$. For both of those functions we can put in any real number, so $D=\{x~|~x\in\mathbb{R}\}$. Looking at the graphs above we see that the maximum function value of both functions is given by $1$ and the minimum by $-1$. This gives us the range $R=\{y~|~-1\le y\le 1\}$.

• Describe how to draw the graph of a function.

  Hints

  Here you see the graph belonging to $f(x)=|x|$.

  The domain is $D=\{x~|~x\in \mathbb{R}\}$ and the range $R=\{y~|~y\ge 0\}$.

  The horizontal axis is called the $x$-axis and the vertical one the $y$-axis.

  Given a function, for each $x$ there exists one $y=f(x)$.

  Solution

  If you have any function you can draw its graph into an $(x,y)$-coordinate system. The $x$-axis is the horizontal axis while the $y$-axis is the vertical axis.

  Each function has a domain, the set of all possible inputs, the $x$-values, and a range with all possible function values, the $y$-values.

  If you'd like to check if a graph could belong to a function you can perform the vertical line test: each line parallel to the $y$-axis has at most one point in common with the graph.

• Examine the domain, range, and asymptotes of the functions.

  Hints

  An asymptote is an invisible line such that a curve, or graph of a function, comes closer and closer to it but will never touch it.

  You can find the range of a function if you know its asymptotes.

  Solution

  $f(x)=\frac{x+1}x=1\frac1x$

  • The denominator cannot equal zero, so $D=\{x~|~x\in\mathbb{R},~x\neq 0\}$.
  • With $f(x)=1+\frac1x$, we can see that $y=1$ is the asymptote.
  • Looking at $1+\frac1x=1$, we can see that $\frac1x=0$, which is not possible. So $R=\{y~|~y\in\mathbb{R},~y\neq 1\}$.

  $g(x)=\frac x{x+1}$

  • The denominator cannot equal zero, so $D=\{x~|~x\in\mathbb{R},~x\neq -1\}$.
  • Putting increasing values (either negative or positive) in for $x$, we see that $y=1$ is the asymptote.
  • Looking at the equation $\frac{x}{x+1}=1$, we can see that $x=x+1$ or $0=1$, which are not possible. So $R=\{y~|~y\in\mathbb{R},~y\neq 1\}$.

  $h(x)=\frac 2x$

  • $D=\{x~|~x\in\mathbb{R},~x\neq 0\}$.
  • The asymptote is $y=0$.
  • $R=\{y~|~y\in\mathbb{R},~y\neq 0\}$.

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

  • The denominator cannot equal zero, so $D=\{x~|~x\in\mathbb{R},~x\neq 1\}$.
  • Putting increasing values (either negative or positive) in for $x$ we see that $y=-1$ is the asymptote.
  • Looking at the equation $\frac{x+2}{-x+1}=-1$, we can see that $x+2=x-1$ or $2=-1$, which are not possible. So $R=\{y~|~y\in\mathbb{R},~y\neq -1\}$.