Question

Show that the function f: R → R given by f(x) = x^3 is injective.

Answer 1

method 1 :- Let x , y ∈ R such that f(x) = f(y)
x³ = y³
taking cube root both sides,
x = y , therefore f is one - one or injective mapping.

method 2 :- if any function f is strictly increasing or strictly decreasing in its domain then, f must be one - one or injective mapping.
it means, if differentiation of f = f' > 0 or f ' < 0 for all belongs to its domain.
then, f is one - one or surjective mapping.

Here, f(x) = x³
differentiate with respect to x
f'(x) = 3x² > 0 for all x belongs to R
then, f is one - one or injective mapping.

Answer 2

Question= that the Signum Function f: R → R, given by

f(x)=⎧⎩⎨⎪⎪⎪⎪⎪⎪10−1for x>0for x=0 is neither one−one nor ontofor x<0

Solution:

Check for one to one function:

For example:

f(0) = 0

f(-1) = -1

f(1) = 1

f(2) = 1

f(3) = 1

Since, for the different elements say f(1), f(2) and f(3), it shows the same image, then the function is not one to one function.

Check for Onto Function:

For the function,f: R →R

f(x)=⎧⎩⎨⎪⎪10−1for x>0for x=0for x<0

In this case, the value of f(x) is defined only if x is 1, 0, -1

For any other real numbers(for example y = 2, y = 100) there is no corresponding element x.

Thus, the function “f” is not onto function.

Hence, the given function “f” is neither one-one nor onto.