Question

Show that the function f : R → R given by f (x) = x³ is injective.

Answer 1

Answer:

Given, f(x)=x³ −1

Let x 1 ,x 2ϵR

Now, f(x 1 )−f(x 2 )

⇒x 1³−1 = x2³−1

⇒x 1³ = x 2³

⇒x 1 = x 2

∴ f(x) is one-one. Also, it is onto as range of f=R.

Hence, it is a bijection.

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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.