# Functions .Evaluate if the function y = 3x^5 + 7x^3 + 2x + 1 is bijective .

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We have to determine if the function y = 3x^5 + 7x^3 + 2x + 1 is bijective.

A function is bijective if it is injective and surjective at the same time. For f(x) to be injective, for no two values of x should the value of f(x) be the same.

y = 3x^5 + 7x^3 + 2x + 1 has odd powers of x, so each value of x gives a distinct value for y. The given function is injective.

The function y = 3x^5 + 7x^3 + 2x + 1 is surjective as every value of the co-domain is given when the function is applied to some value in the domain.

**As y = 3x^5 + 7x^3 + 2x + 1 is injective and surjective, it is bijective.**

To prove that a function is bijective, we have to verify if the function is one to one and on-to.

To verify if the function is one to one, we'll calculate the first derivative and we'll check if it's positive or negative.

In our case, the derivative of the function is:

f'(x)=15x^4 + 21x^2 +2

In order to analyze the monotony of the function, we'll calculate the roots of the first derivative, which in this case is a quadratic function.

We'll make the substitution of x^2=t

The equation of the first derivative becomes:

15t^2+21t+2=0

Delta=(21)^2 - 4*15*2>0

delta = 441 - 120

delta = 321

sqrt delta = 17.9

t1=(-21+17)/30<0

t2=(-21-17)/30<0

So,

x^2=-t1

x^2+t1>0 and

x^2+t2>0, so the function is positive for any value of x.

If the function is positive for any value of x, that means that is a strictly increasing function on domain of definition of function.

Any function strictly increasing on it's domain, is an one to one function!

For any x1<x2 => f(x1)<f(x2), which is the rule of an injective function!

lim f(x)=-infinity, when x tends to -infinity and

lim f(x)=infinity,when x tends to infinity .

From these limits, it is obvious that the function is increasing continuously.

If the function is continuous, then it is an on-to function.

If the function is one to one and on-to same time, that means that the function is bijective!