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

a) $f(x)=-3x+4$

b) $f\left(x\right)=-3{x}^{2}+7$

c) $f(x)=\frac{x+1}{x+2}$

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

Joni Kenny
2021-09-07
Answered

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

a) $f(x)=-3x+4$

b) $f\left(x\right)=-3{x}^{2}+7$

c) $f(x)=\frac{x+1}{x+2}$

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

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Ayesha Gomez

Answered 2021-09-08
Author has **104** answers

Part a-

$f(x)=-3x+4$ is a bijection.

It is one-to-one as $f\left(x\right)=f\left(y\right)\Rightarrow -3x+4=-3y+4\Rightarrow x=y$

It is onto as $f\left(\frac{4-x}{3}\right)=x$

Part b-

$f(-x)=f(x)$.

The function is not a bijection as a result.

Part c-

No real number exists that is such that $f\left(x\right)=\frac{x+1}{x+2}=1$. Hence the function is not a bijection.

Part d-

$f\left(x\right)={x}^{5}+1$ is a bijection.

It is a strictly increasing function.

Result:

Part a - Yes

Part b - NO

Part c - No

Part d - Yes

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