Question 332816
define the function F on [-7,-1] by F(x)=x^2+8

Determine whether F is one to one
<pre><b>
First let's draw the graph and see if it looks like it is

one-to-one, then we'll prove whether it is or not

{{{drawing(300,300,-9,2,-3,64,

line(-7,-10,-7,62), line(-1,-10,-1,62),

graph(300,300,-9,2,-3,64,(x^2+8)*

(sqrt(-1-x)/sqrt(-1-x))


*(sqrt(x+7)/sqrt(x+7)))   )}}}

It looks one-to-one because it does not pass beside itself

at any point.

Let's pass some horizontal lines through it, and see if

they all pass through the graph only once.

{{{drawing(300,300,-9,2,-3,64,

line(-7,-10,-7,62), line(-1,-10,-1,62),

green( line(-10,15,3,15),line(-10,21,3,21),line(-10,27,3,27),
line(-10,33,3,33),line(-10,39,3,39),line(-10,45,3,45),line(-10,51,3,51)),

graph(300,300,-9,2,-3,64,(x^2+8)*

(sqrt(-1-x)/sqrt(-1-x))


*(sqrt(x+7)/sqrt(x+7)))   )}}}

So it looks like it passes the horizontal line test, so we 

believe that it is one-to-one.  However "looking and seeing" does not 

prove anything.  So let's prove that it is one-to-one.

PROOF:

Suppose, for contradiction, that it is not one-to one.

Then there exists two numbers a and b, both in the domain of 

F(x), which is [-7.-1], meaning that they are both negative 

since the domain contains only negative numbers, such that

          F(a) = F(b)

Then

        a² + 8 = b² + 8

therefore

       a² - b² = 0

(a - b)(a + b) = 0

a - b = 0  or  a + b = 0

    a = b  

If a = b then that contradicts the assumption that a and b
are different numbers.

The other equation a + b = 0 is not possible because a and b
are both negative numbers (since the domain [-7,1] contains 
only negative numbers, and the sum of two negative numbers is
always negative, and never 0.  Therefore we have proved that
the function is one-to-one.


The range is [f(-1), f(-7)] = [(-1)²+8, (-7)²+8] = [1+8,49+8] = [9,57]


Edwin</pre>