Question 639145
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^2\ -\ 4]


Solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 4\ =\ 0]


for


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1,\ x_2]


The *[tex \LARGE x]-intercept<b><i>s</i></b> are then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x_1,0)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x_2,0)]


Evaluate *[tex \LARGE y_i\ =\ f(0)], then *[tex \LARGE (0,y_i)] is the *[tex \LARGE y]-intercept.


NOTE: These are NOT the answers.  You have to calculate the actual values of *[tex \LARGE x_1,\ x_2] and *[tex \LARGE y_i]


This is a polynomial function with real coefficients, therefore the domain is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{dom}\, f\ =\ \left{x\ \in\ \mathbb{R}\right}]


To define the range and intervals of increase and decrease, we first need to calculate the function maximum.


Take the first derivative using the power rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d\left(f(x)\right)}{dx}\ =\ 2x]


Set the first derivative equal to zero and solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ 0\ \Rightarrow\ x\ =\ 0]


Hence a local extremum exists at *[tex \LARGE x\ =\ 0]


Take the second derivative


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2\left(f(x)\right)}{dx^2}\ =\ 2]


Which is positive *[tex \LARGE \forall\ x\ \in\ \text{dom}\, f], therefore the extremum at *[tex \LARGE x\ =\ 0] is a local minimum, and the minimum value is *[tex \LARGE f(0)\ =\ -4]


Since this is a polynomial function it is continuous over the entire domain, hence the range is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{ran}\, f\ =\ \left{y\ \in\ \mathbb{R}\ |\ y\ \geq\ -4\right}]


The first derivative has a single root, namely *[tex \LARGE x\ =\ 0], hence there are two intervals to consider:  *[tex \LARGE \left(-\infty,\,0\right)] and *[tex \LARGE \left(0,\,\infty\right)].


Choose a value in the first interval, say -1.  *[tex \LARGE f'(-1)\ =\ -2], hence the interval *[tex \LARGE \left(-\infty,\,0\right)] is an interval of decrease.


Choose a value in the second interval, say 1.  *[tex \LARGE f'(1)\ =\ 2], hence the interval *[tex \LARGE \left(0,\,\infty\right)] is an interval of increase.


This is an even-degree polynomial with a positive lead coefficient, hence both ends take off to positive infinity.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>