Question

The Hyperbolic Sine (sinh(x)) and Hyperbolic Cosine (cosh(x)) functions are defined as such: sin h(x) = e^x - e^-x/2 cosh(x) = e^x + e^-x/2 Compute the y-Intercept of both functions Then Determine the end behavior of these two functions by analyzing the limit as x goes to positive infinity and as x goes to negative infinity

Answer 1

Answer:

y-incercepts:

sinh(x):0, cosh(x)=1

Limits:

positive infinity: sinh(x): infinity, cosh(x): infinity

negative infinity: sinh(x): - infinity, cosh(x): infinity

Step-by-step explanation:

We are given that

$\sinh(x)=\frac{e^{x}-e^{-x}}{2}$

$\cosh(x)=\frac{e^{x}+e^{-x}}{2}$

To find out the y-incerpt of a function, we just need to replace x by 0. Recall that $e^{0}=1$. Then,

$\sinh(0) = \frac{1-1}{2}=0$

$\cosh(0) = \frac{1+1}{2}=1$

For the end behavior, recall the following:

$\lim_{x\to \infty}e^{x} = \infty, \lim_{x\to \infty}e^{-x} = 0$

$\lim_{x\to -\infty}e^{x} = 0, \lim_{x\to -\infty}e^{-x} = \infty$

Using the properties of limits, we have that

$\lim_{x\to \infty} \sinh(x) =\frac{1}{2}(\lim_{x\to \infty}e^{x}-\lim_{x\to \infty}e^{-x})=(\infty -0) = \infty$

$\lim_{x\to \infty} \cosh(x) =\frac{1}{2}(\lim_{x\to \infty}e^{x}+\lim_{x\to \infty}e^{-x}) =(\infty -0)= \infty$

$\lim_{x\to -\infty} \sinh(x) =\frac{1}{2}(\lim_{x\to -\infty}e^{x}-\lim_{x\to -\infty}e^{-x}) = (0-\infty)=-\infty$

$\lim_{x\to -\infty} \cosh(x) =\frac{1}{2}(\lim_{x\to -\infty}e^{x}+\lim_{x\to -\infty}e^{-x}) =(0+\infty)= \infty$