Question

1.Find the Domain and Range of the following functions if 𝑓: 𝑅 → 𝑅 i) 𝑓(𝑥) = 12 + 𝑙𝑜𝑔 (1 − 𝑥^2 ) ii) 𝑓(𝑥) = 𝑥^2 +1/𝑥^2 +5

2. Find if the functions are surjective, injective and bijective if 𝑓: 𝑅 → 𝑅 i) 𝑓(𝑥) = 𝑥^5 + 5 ii) 𝑓(𝑥) = 𝑒^4𝑥 

Answer 1

1.

i)

$$f(x)=12+\log(1-x^2)$$

$$1-x^2>0=>-1<x<1$$

Domain: $(-1, 1)$

$$0<1-x^2\le1$$

Range: $(-\infin, 12].$

ii)

$$f(x)=x^2+\dfrac{1}{x^2}+5$$

$$x\not=0$$

Domain: $(-\infin, 0)\cup (0, \infin)$

$$x^2+\dfrac{1}{x^2}\ge2, x\in\R$$

Range: $[7, \infin).$

2)

i)

$$f(x)=x^5+5$$

$f(x)$ is strictly increasing on $\R.$ So, if $n, m\in \R, n\not=m,$ then $n>m$ or $n<m.$

Without loss of generality, assume that $n>m.$

Then we have $n^5+5>m^5+5.$ So if $n\not=m,$ then $f(n)\not=f(m).$

Therefore the function $f(x)=x^5+5$ is injective.

$f(x)$ is strictly increasing on $\R.$ For every $\forall y\in \R, \exist x\in \R$ such that

$$f(x)=y$$

$$x^5+5=y=>x=(y-5)^{1/5}$$

Therefore the function $f(x)=x^5+5$ is surjective.

$f(x)$ is strictly increasing on $\R.$ For every $\forall y\in \R, \exist! x\in \R$ such that

$$f(x)=y$$

$$x^5+5=y=>x=(y-5)^{1/5}$$

Therefore the function $f(x)=x^5+5$ is bijective.

ii)

$$𝑓(𝑥) = 𝑒^{4𝑥 }$$

$f(x)$ is strictly increasing on $\R.$ So, if $n, m\in \R, n\not=m,$ then $n>m$ or $n<m.$

Without loss of generality, assume that $n>m.$

Then we have $e^{4n}>e^{4m}$ . So if $n\not=m,$ then $f(n)\not=f(m).$

Therefore the function $f(x)=e^{4x}$ is injective.

If $y\le0,$ we cannot find $x\in \R$ such that $f(x)=y.$

Therefore the function $f(x)=e^{4x}$ is not surjective and is not bijective.