Prove that the funcion f RR defined by fx x5 7 is onetooneS

Prove that the funcion f: R->R defined by f(x) = x^5 +7 is one-to-one.

f(x) = x^5 + 7

Let us consider two values x = a and x = b :

f(a) = a^5 + 7
f(b) = b^5 + 7

Now, for the function f(x) to be one to one, it would indicate that for any \'a\' and \'b\', f(a) and f(b) are different values....

Now, let us assume that f(a) = f(b)...
a^5 + 7 = b^5 + 7

Subtract 7 both sides :

a^5 = b^5

Fifthroot both sides :

a = b

Since f(a) = f(b) directly implies that a = b, it has been proven that the function f(x) = x^5 + 7 is 1-to-1