Show that the function f : R → R : f (x) = x4 is neither one-one nor onto.
To prove: function is neither one-one nor onto
Given: f : R → R : f (x) = x4
We have,
f(x) = x4
For, f(x1) = f(x2)
⇒ x14 = x24
⇒ (x14 - x24) = 0
⇒(x12 - x22) (x12 + x22) = 0
⇒ (x1 - x2) (x1 + x2) (x12 + x22) = 0
⇒ x1 = x2 or, x1 = -x2 or, x12 = -x22
We are getting more than one value of x1 (no unique image)
∴ f(x) is not one-one
f(x) = x4
Let f(x) = y such that ![]()
⇒ y = x4
![]()
If y = -2, as ![]()
Then x will be undefined as we can’t place the negative value under the square root
Hence f(x) is not onto
Hence Proved
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