Find out if the following functions are one-to-one and/or onto. a. 𝑓: 𝑍 → 𝑅, 𝑓(𝑥) = 𝑥 3 + 1 b. 𝑓: 𝑅 + → 𝑅 +, 𝑓(𝑥) = |𝑥| + 5
a:f−one−to−one:f(x)=f(y)⇒x3+1=y3+1⇒x=yf−not onto:f(x)=3⇒x3+1=3⇒x3=2⇒x∉Zb:f−one−to−one:f(x)=f(y)⇒∣x∣+5=∣y∣+5⇒∣x∣=∣y∣⇒[x,y∈R+]⇒x=yf−not onto:f(x)=1⇒x+5=1⇒x=−4⇒x∉R+a:\\f-one-to-one:\\f\left( x \right) =f\left( y \right) \Rightarrow x^3+1=y^3+1\Rightarrow x=y\\f-not\,\,onto:\\f\left( x \right) =3\Rightarrow x^3+1=3\Rightarrow x^3=2\Rightarrow x\notin \mathbb{Z} \\b:\\f-one-to-one:\\f\left( x \right) =f\left( y \right) \Rightarrow \left| x \right|+5=\left| y \right|+5\Rightarrow \left| x \right|=\left| y \right|\Rightarrow \left[ x,y\in \mathbb{R} _+ \right] \Rightarrow x=y\\f-not\,\,onto:\\f\left( x \right) =1\Rightarrow x+5=1\Rightarrow x=-4\Rightarrow x\notin \mathbb{R} _+a:f−one−to−one:f(x)=f(y)⇒x3+1=y3+1⇒x=yf−notonto:f(x)=3⇒x3+1=3⇒x3=2⇒x∈/Zb:f−one−to−one:f(x)=f(y)⇒∣x∣+5=∣y∣+5⇒∣x∣=∣y∣⇒[x,y∈R+]⇒x=yf−notonto:f(x)=1⇒x+5=1⇒x=−4⇒x∈/R+
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