How do I show that this function is one-to-one algebraically?
B3%7D%20%2B8" id="TexFormula1" title="f(x)= (x-2)^{3} +8" alt="f(x)= (x-2)^{3} +8" align="absmiddle" class="latex-formula">
1 answer:
For a one-to-one function, f(x) = f(y).
So, lets find f(x) and f(y).
We know,
f(x) = (x - 2)³ + 8
Now,
f(y) = (y - 2)³ + 8
Now, we said earlier, for a function to be one-to-one, f(x) = f(y).
Therefore,
f(x) = f(y)
(x - 2)³ + 8 = (y - 2)³ + 8
(x - 2)³ = (y - 2)³
x - 2 = y - 2
x = y
Since we got x = y, we know for every x, there is one and only one y. Therefore, the function is one-to-one function.
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