Show that the exponential function f: R → R, given by f(x) = e x , is one – one but not onto. What happens if the co – domain is replaced by R 0 + (set of all positive real numbers).

Question

Philoid · Accepted Answer

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each, there exists such that f(a) = b

Now, given by f(x) = e^x

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

So, from definition

⇒ f(x) = f(y)

⇒ e^x = e^y

⇒

⇒ e^{x – y} = 1

⇒ e^{x – y} = e⁰

⇒ x – y = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Here range of f = (0,∞) ≠ R

Therefore f is not onto

Now if co – domain is replaced by R₀ ⁺ (set of all positive real numbers) i.e (0,∞) then f becomes an onto function.

Show that the exponential function f: R → R, given by f(x) = ex, is one – one but not onto. What happens if the co – domain is replaced by R0 + (set of all positive real numbers).

Show that the exponential function f: R → R, given by f(x) = e^x, is one – one but not onto. What happens if the co – domain is replaced by R₀ ⁺ (set of all positive real numbers).