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

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### 2 Answers

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! **

y = 3x^5 + 7x^3 + 2x + 1.

This function y = 3x^5 + 7x^3 + 2x + 1 is continous differentiable function defined for all x from -infinity to infinity.

We shall now verify whether the function is increasing or decreasing by differentiating the function.

The first derivative is y' = (3x^5+7x^3 +2+1)' = (15x^4+7x^2+2) > =1 for all x since each term is nonnegative and the constant term is 1.

Therefore y = 3x^5 + 7x^3 + 2x + 1 is an increasing function for all x. So this function is bijective as all monotonously increasing fuctions are bijective .