How to determine whether f(x)= 2x²-4x -1 is one to one function given that x is more than or equal to 1?

3 Answers
Jul 11, 2015

Answer:

#d/(dx) f(x) = 4x-4 = 4(x-1) >= 0# when #x >= 1# and is only 0 when #x=1#.

So #f(x)# is strictly monotonically increasing for #x in [1, oo)# so is 1-1

Explanation:

This is a parabola with vertex at #x=1# so is strictly monotonic for #x in [1, oo)# so one to one.

Jul 11, 2015

Answer:

If #x_1, x_2 in [1, oo)# show that #f(x_1) = f(x_2) => x_1 = x_2#

by directly solving. So #f(x)# is 1-1 on #[1, oo)#

Explanation:

#f(x) = 2x^2-4x-1 = 2(x-1)^2-3#

So if #f(x_1) = f(x_2)# then

#2(x_1-1)^2-3 = 2(x_2-1)^2-3#

Add #3# to both sides to get:

#2(x_1-1)^2 = 2(x_2-1)^2#

Divide both sides by #2# to get:

#(x_1-1)^2 = (x_2-1)^2#

So:

#abs(x_1-1) = abs(x_2-1)#

Given #x_1 >= 1# and #x_2 >= 1# then

#x_1 - 1 >= 0# so #abs(x_1 - 1) = x_1 - 1#

#x_2 - 1 >= 0# so #abs(x_2 - 1) = x_2 - 1#

So

#x_1 - 1 = abs(x_1 - 1) = abs(x_2 - 1) = x_2 - 1#

Add #1# to both ends to get:

#x_1 = x_2#

So #f(x_1) = f(x_2) => x_1 = x_2# for all #x_1, x_2 in [1, oo)#

Jul 11, 2015

Answer:

This can also be determined graphically.

Explanation:

The graph of #f(x) = 2x^2-4x-1# is a parabola with vertex: #(1, -3)#

(Use the vertex formula or put the expression in vertex form: #f(x) = 2(x-1)^2 -3#.)

So with no restriction on the domain, the graph looks like:

graph{2x^2-4x-1 [-6.97, 10.81, -5.69, 3.2]}

With its natural domain for the expression, the function #g(x) = 2x^2-4x-1# (domain #(-oo, oo)#) fails the horizontal line test. It is not one-to-one.

When we restrict the domain to #x > = 1#, we eliminate the "left half" of the parabola,

The graph of #f(x)# looks like:

graph{y = (2x^2-4x-1)*sqrt(x-1)/sqrt(x-1) [-4.32, 8.17, -4.367, 1.88]}

This function passes the horizontal line test. It is one-to-one.