 2023-03-30

Whether each of these functions is a bijection from R to R.

a) $f\left(x\right)=-3x+4$

b) $f\left(x\right)=-3{x}^{2}+7$

c) $f\left(x\right)=\frac{x+1}{x+2}$
$d\right)f\left(x\right)={x}^{5}+1$

? Kieran Orozco

To determine whether each of the given functions is a bijection from $ℝ$ to $ℝ$, we need to check if they satisfy both injectivity (one-to-one) and surjectivity (onto) properties.
a) $f\left(x\right)=-3x+4$
To check if this function is a bijection, we need to verify if it is both one-to-one and onto.
Injectivity:
Let's assume that $f\left({x}_{1}\right)=f\left({x}_{2}\right)$ for some ${x}_{1}$ and ${x}_{2}$ in $ℝ$. Then we have $-3{x}_{1}+4=-3{x}_{2}+4$. By simplifying the equation, we find ${x}_{1}={x}_{2}$. This shows that different inputs map to different outputs, and therefore $f\left(x\right)$ is injective.
Surjectivity:
To check surjectivity, we need to determine if every real number has a corresponding input value. In this case, since $f\left(x\right)$ is a linear function with a negative coefficient on $x$, the function will not cover the entire range of $ℝ$. It will only cover a subset of $ℝ$. Therefore, $f\left(x\right)$ is not surjective.
In summary, $f\left(x\right)=-3x+4$ is injective but not surjective, so it is not a bijection.
b) $f\left(x\right)=-3{x}^{2}+7$
For this function, let's check its injectivity and surjectivity.
Injectivity:
If we assume $f\left({x}_{1}\right)=f\left({x}_{2}\right)$, then $-3{x}_{1}^{2}+7=-3{x}_{2}^{2}+7$. By simplifying the equation, we have ${x}_{1}^{2}={x}_{2}^{2}$. However, this does not guarantee that ${x}_{1}={x}_{2}$, as there can be different input values with the same square. Therefore, $f\left(x\right)$ is not injective.
Surjectivity:
To check surjectivity, we need to determine if every real number has a corresponding input value. However, since the function is a quadratic with a negative coefficient on ${x}^{2}$, it will only cover a subset of $ℝ$. Thus, $f\left(x\right)$ is not surjective.
In summary, $f\left(x\right)=-3{x}^{2}+7$ is neither injective nor surjective, so it is not a bijection.
c) $f\left(x\right)=\frac{x+1}{x+2}$
Let's examine the injectivity and surjectivity of this function.
Injectivity:
Assuming $f\left({x}_{1}\right)=f\left({x}_{2}\right)$, we have $\frac{{x}_{1}+1}{{x}_{1}+2}=\frac{{x}_{2}+1}{{x}_{2}+2}$. By cross-multiplication and simplification, we find ${x}_{1}={x}_{2}$. This demonstrates that $f\left(x\right)$ is injective.
Surjectivity:
For every value of $y$ in the range, we can find a corresponding $x$ such that $f\left(x\right)=y$. Therefore, $f\left(x\right)$ is surjective.
In conclusion, $f\left(x\right)=\frac{x+1}{x+2}$ is both injective and surjective, making it a bijection.
d) $f\left(x\right)={x}^{5}+1$
Let's analyze the injectivity and surjectivity of this function.
Injectivity:
Assume $f\left({x}_{1}\right)=f
\left({x}_{2}\right)$
, which gives us ${x}_{1}^{5}+1={x}_{2}^{5}+1$. By subtracting 1 from both sides and simplifying, we find ${x}_{1}^{5}={x}_{2}^{5}$. Since the function is a monomial with an odd power, we can conclude that ${x}_{1}={x}_{2}$. Thus, $f\left(x\right)$ is injective.
Surjectivity:
For every value of $y$ in the range, we can find a corresponding $x$ such that $f\left(x\right)=y$. Hence, $f\left(x\right)$ is surjective.
In summary, $f\left(x\right)={x}^{5}+1$ is both injective and surjective, indicating that it is a bijection.

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