In this lesson we’ll learn how to draw graphs of hyperbolic functions.

#### Hyperbolic sine, $sinh$

As we learned before, hyperbolic sine, $sinh:\mathbb{R} \rightarrow \mathbb{R}$, is monotonically increasing function defined with

$$\displaystyle{sinhx=\frac{e^{x}-e^{-x}}{2}}.$$

What would its graph look like? Maybe we wouldn’t know how to draw it at the first glance, but we have to remember its definition. Function $sinhx$ is defined with exponentional functions and we know what their graphs look like. Lets use that knowledge to draw $sinhx$.

First, lets start with calculating the value of $sin0$.

**For $x=0$, **$e^{x}=1$ and $e^{-x}=1$. As a result,

$$\displaystyle{sinh0=\frac{e^{0}-e^{-0}}{2}=\frac{1-1}{2}=0}.$$

To see how the rest of the hyperbolic sine behaves, we have to recall the graphs of two exponential functions. Rewrite $sinh$ as $$\displaystyle{sinhx=\frac{e^{x}}{2} – \frac{e^{-x}}{2}}$$ and lets sketch those graphs.

**For positive $x$
**As $x$ gets larger, $e^{x}$ quickly increases, but $e^{-x}$ qucikly decreases. Consequently, $\displaystyle{-\frac{e^{-x}}{2}}$ gets very small. Therefore, as $x$ gets larger, $sinhx$ gets closer and closer to $\displaystyle{\frac{e^{x}}{2}}$,

$sinhx \approx \displaystyle{\frac{e^{x}}{2}}$, for large $x$.

Even though $\displaystyle{- \frac{e^{-x}}{2}}$ gets really small, it is always less than zero, meaning the graph of $sinhx$ will always be below the graph of $\displaystyle{\frac{e^{x}}{2}}$. As $x$ gets larger, the difference between two graphs, graphs of $sinhx$ and $\displaystyle{\frac{e^{x}}{2}}$, gets smaller and smaller.

**For negative $x$
**As $x$ gets more negative, $e^{-x}$ quickly becomes larger and negative, and $e^{x}$ quickly decreases. Consequently, $\displaystyle{\frac{e^{x}}{2}}$ get very smaller. Therefore, $sinhx$ gets closer and closer to $\displaystyle{-\frac{e^{-x}}{2}}$,

$sinhx \approx \displaystyle{-\frac{e^{-x}}{2}}$, for large negative $x$.

The graph of $sinhx$ will always stay above the $\displaystyle{-\frac{e^{-x}}{2}}$ when $x$ is negative because, even though $\displaystyle{\frac{e^{x}}{2}}$ gets very small, it’s always greater than zero. As $x$ gets more and more negative the difference between graphs of $sinhx$ and $\displaystyle{-\frac{e^{-x}}{2}}$ gets smaller and smaller.

Finally, lets sketch the graph of $sinhx$ between the graphs of $\displaystyle{\frac{e^{x}}{2}}$ and $\displaystyle{\frac{e^{-x}}{2}}$

From graph we can clearly see that hyperbolic sine is an odd function, $sinh(-x) = – sinhx$.

#### Hyperbolic cosine, $cosh$

Hyperbolic cosine is a function $cosh:\mathbb{R} \rightarrow [1, +\infty)$ defined with

$$\displaystyle{coshx=\frac{e^{x}+e^{-x}}{2}}.$$

We’ll use the knowledge of graphs of $e^{x}$ and $e^{-x}$ to draw a graph of $coshx$. First, let’s calculate the value of $cosh0$.

For $x=0$, $e^{x}=1$ and $e^{-x}=1$. As a result,

$$\displaystyle{cosh0=\frac{e^{0}+e^{-0}}{2}=\frac{1+1}{2}=1}.$$

To see how the rest of the hyperbolic cosine behaves, we have to recall the graphs of two exponential functions. Rewrite $cosh$ as

$$\displaystyle{cosx=\frac{e^{x}}{2} + \frac{e^{-x}}{2}}$$

and lets sketch those graphs.

**For positive $x$
**As $x$ gets larger, we can see that $e^x$ quickly increases, but $e^{-x}$ quickly decreases. As a result, $\displaystyle{\frac{e^{-x}}{2}} $ gets very small. Therefore, as $x$ gets larger, $coshx$ gets closer and closer to $\displaystyle{\frac{e^{x}}{2}} $,

$coshx \approx \displaystyle{\frac{e^{x}}{2}}$, for large $x$.

Even though $\displaystyle{\frac{e^{-x}}{2}}$ gets very small, it is always greater than zero, meaning the graph of $coshx$ will always be above the graph of $\displaystyle{\frac{e^{x}}{2}}$. As $x$ gets larger, the difference between two graphs, graphs of $cosx$ and $\displaystyle{\frac{e^{x}}{2}}$, gets smaller and smaller.

**For negative $x$
**As $x$ becomes more negative, $e^{-x}$ quickyl increases, but $e^{x}$ quickly decreases. Consequently, $\displaystyle{\frac{e^{x}}{2}} $ gets very small. Therefore, as $x$ get more negative, $coshx$ gets closer and closer to $\displaystyle{\frac{e^{-x}}{2}} $,

$coshx \approx \displaystyle{\frac{e^{-x}}{2}}$, for large negative $x$.

Even though $\displaystyle{\frac{e^{x}}{2}}$ gets very small, it is always greater than zero, meaning the graph of $coshx$ will always be above the graph of $\displaystyle{\frac{e^{-x}}{2}}$. As $x$ gets larger, the difference between two graphs, graphs of $cosx$ and $\displaystyle{\frac{e^{x}}{2}}$, gets smaller and smaller.

Finally, we can sketch the graph of $coshx$.

In contrast to hyperbolic sine, $cosh$ is an even function, $cosh(-x)=cosh(x)$. Because of that, graph is symmetric with respect to the $y$-axis.

#### Hyperbolic tangent, $tanh$

Hyperbolic tangent, $tanh: \mathbb{R} \rightarrow (-1,1)$, is monotonically increasing fucntion defined with

$$\displaystyle{tanhx=\frac{sinhx}{coshx}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}}$$

Since we can write $tanhx$ in terms of $sinh$ and $cosh$, we will use the knowledge of these two functions to sketch $tanhx$.

Firstly, for better understanding of the relationship between $sinhx$ and $coshx$, lets sketch them on the same graphs.

Then, let’s see the value of $tanhx$ for $x=0$.

$$\displaystyle{tanh0=\frac{sinh0}{cosh0}=\frac{0}{1}}=0$$

**For positive $x$
**As $x$ gets larger, $sinhx$ gets closer to $coshx$, meaning $sinhx \approx coshx$. When we apply that to the equation above, we can see that $tanhx$ gets closer and closer to $1$,

$tanhx \approx 1$, for large $x$

Function of $sinhx$ is always less than $coshx$, so $tanhx$ is always less than $1$. It gets closer to it as $x$ gets larger, but never reaches it.

**For negative $x$
**Similarly as before, as $x$ gets more and more negative, $sinh \approx -coshx$. Put that into definition of $tanhx$, we can see

$tanhx \approx -1$, for large negative $x$.

Since $sinhx$ is always greater than $-coshx$, $tanhx$ is always greater than $-1$. It gets closer to it as $x$ gets larger, but never reaches it.

Finally, lets sketch the graph of $tanhx$.

Notice that $tanhx$ is also an odd function, $tanh(-x)=-tanh(x)$.

#### Hyperbolic cotangent, $coth$

Hyperbolic cotangent is a function $coth: \mathbb{R} \backslash \{0\} \rightarrow (-\infty,-1) \cup (1,+\infty)$ defined with

$$\displaystyle{cothx=\frac{coshx}{sinhx}=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}}$$

To make the sketching easier, we’ll try to write the fraction a little bit different:

$$\displaystyle{cothx=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}= \frac{e^{x}-e^{-x}+2e^{-x}}{e^{x}-e^{-x}}= 1+ \frac{2e^{-x}}{e^{x}-e^{-x}}}$$

Now let’s evaluate this new representation of $cothx$.

**For $x=0$**, $\displaystyle{coth0= 1+ \frac{2e^{-0}}{e^{0}-e^{-0}} = \frac{2}{1-1}}$ is undefined! As a result, we should have a vertical asymptotic line at $x=0$.

**For positive $x$
**What happens when $x$ approaches $+\infty$? Lets take a look at fraction $\displaystyle{\frac{2e^{-x}}{e^{x}-e^{-x}}}$. For $x \rightarrow +\infty $ the numerator $2e^{-x}$approaches $0$. As a result, the whole fraction goes to $0$. Now just add the remaining part of $cothx$ representation, $1+0$ and we have that $\displaystyle{1+ \frac{2e^{-x}}{e^{x}-e^{-x}}} \rightarrow 1$, as $x$ approaches $+\infty$.

However, $cothx$ never reaches $1$, which leaves us with asymptotic line for $y=1$.

**For negative $x$
**As $x$ approaches $-\infty$ the whole fraction $\displaystyle{1+\frac{2e^{-x}}{e^{x}-e^{-x}}}$ goes to $-1$. Why is that?

As $x$ gets more and more negative, fraction $\displaystyle{\frac{2e^{-x}}{e^{x}-e^{-x}}}$ approaches $-2$. When we add $1+(-2)$ we get that $\displaystyle{1+\frac{2e^{-x}}{e^{x}-e^{-x}}} \rightarrow -1$, for $x \rightarrow -\infty$.

However, $cothx$ never reaches $-1$, which leaves us with asymptotic line for $y=-1$.

Now we know enough to sketch the graph. It would look like this:

As we can see, hyperbolic cotangent is an odd function, meaning $coth(-x)=-coth(x)$.

And finally, we will briefly mention hyperbolic secant and cosecant.

#### Hyperbolic secant, $sech$

Hyperbolic secant is a function $sech: \mathbb{R} \rightarrow (0,1) $ defined with

$$\displaystyle{sechx=\frac{1}{coshx}=\frac{2}{e^{x}+e^{-x}}}$$

Graph looks like this:

The second even function in this family of functions is hyperbolic secant, $sech(-x)=sech(x)$.

That’s because it’s directly connected to $coshx$.

#### Hyperbolic cosecant, $csch$ or $cosech$

Hyperbolic cosecant is a function $csch: (-\infty,-1) \cup (1,+\infty) \rightarrow \mathbb{R} \backslash \{0\}$

$$\displaystyle{cschx=\frac{1}{sinhx}=\frac{2}{e^{x}-e^{-x}}}$$

As we can see from the graph, it’s an odd function, meaning $csch(-x)=-csch(x)$