Determine whether each of these functions is a bijection from R to R. a) f (x)

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) $f(x)=-3x+4$
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 $f(x_1)=f(x_2)$ for some $x_1$ and $x_2$ in $ℝ$. Then we have $-3x_1+4=-3x_2+4$. By simplifying the equation, we find $x_1=x_2$. This shows that different inputs map to different outputs, and therefore $f(x)$ is injective.
Surjectivity:
To check surjectivity, we need to determine if every real number has a corresponding input value. In this case, since $f(x)$ 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, $f(x)$ is not surjective.
In summary, $f(x)=-3x+4$ is injective but not surjective, so it is not a bijection.
b) $f(x)=-3x^2+7$
For this function, let's check its injectivity and surjectivity.
Injectivity:
If we assume $f(x_1)=f(x_2)$, then $-3x_1^2+7=-3x_2^2+7$. 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, $f(x)$ 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, $f(x)$ is not surjective.
In summary, $f(x)=-3x^2+7$ is neither injective nor surjective, so it is not a bijection.
c) $f(x)=\frac{x+1}{x+2}$
Let's examine the injectivity and surjectivity of this function.
Injectivity:
Assuming $f(x_1)=f(x_2)$, 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 $f(x)$ is injective.
Surjectivity:
For every value of $y$ in the range, we can find a corresponding $x$ such that $f(x)=y$. Therefore, $f(x)$ is surjective.
In conclusion, $f(x)=\frac{x+1}{x+2}$ is both injective and surjective, making it a bijection.
d) $f(x)=x^5+1$
Let's analyze the injectivity and surjectivity of this function.
Injectivity:
Assume $f(x_1)=f<br>(x_2)$, 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, $f(x)$ is injective.
Surjectivity:
For every value of $y$ in the range, we can find a corresponding $x$ such that $f(x)=y$. Hence, $f(x)$ is surjective.
In summary, $f(x)=x^5+1$ is both injective and surjective, indicating that it is a bijection.