Question

Show that this function is one-to-one algebraically.
f(x)=(x−2)^3+8

Answer 1

@lexxii, First replace f(x) with y

Answer 2

Can you do that?

Answer 3

y=(x-2)^3+8

Answer 4

Now swap the x and the y.

Answer 5

x=(y-2)^3+8

Answer 6

Subtract 8 from both sides

Answer 7

x-8=(y-2)^3+8-8
x-8=(y-2)^3 ?

Answer 8

Very good. Now apply the cube root to both sides.

Answer 9

Would the cube root go to the x or?

Answer 10

It goes over the entire expression on either side.

Answer 11

so (x-8)^3=(y-2)^3 ?

Answer 12

Well, you "cubed" one side and left the other sided "cubed". but you did not apply the "cube root" to both sides.

Answer 13

I recommend you either use \(\LaTeX\) or the drawing feature to write your steps.

Answer 14

?

Answer 15

Almost. We want to eliminate the cubed term on the RHS.

Answer 16

To do this we apply the cube root. So it should be
\(\sqrt[3]{x - 8} = \sqrt[3]{(y - 2)^3}\)

Answer 17

When done in this manner, the cube and cube root cancel on the RHS

Answer 18

oh okay

Answer 19

Actually we're finding the inverse of the function using this method. It seems that to determine whether the function is one-to one, we have to first re-write the function in terms of f(a) and f(b) then set the expressions equal to each other.

Answer 20

So what we should do is:
\(f(a) = (a - 2)^3 + 8\)
\(f(b) = (b - 2)^3 + 8\)

Then set \(f(a) = f(b)\) to get

\((a - 2)^3 + 8 = (b - 2)^3 + 8\)
And then keep applying inverse operations until we get \(a = b\)

Answer 21

Okay that makes sense I can finish it on my own from there thank you so much!

Answer 22

You're welcome