define the function F on [-7,-1] by F(x)=x^2+8
Determine whether F is one to one
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
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.
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
