a) Consider the following functional relation, f, defined as:f:R→R,f(x)=x2Determine whether or not fis a bijection. If it is, prove it. If it is not, show why it is not&nbsp;b) Consider the set&nbsp;F={y∣y=ax3+b}a, b being constants such that&nbsp;a≠0&nbsp;and&nbsp;x∈R.&nbsp;Is&nbsp;F=R? If so, prove it. If not, explain in details why it is not the case.

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a) Consider the following functional relation, f, defined as:f:R→R,f(x)=x2Determine whether or not fis a bijection. If it is, prove it. If it is not, show why it is not&amp;nbsp;b) Consider the set&amp;nbsp;F={y∣y=ax3+b}a, b being constants such that&amp;nbsp;a≠0&amp;nbsp;and&amp;nbsp;x∈R.&amp;nbsp;Is&amp;nbsp;F=R? If so, prove it. If not, explain in details why it is not the case.

obduciramwz6 · Accepted Answer

Step 1 (a). Given: f:R→R,f(x)=x2To check:f is one - one:f(x)=f(y)⇒x=yHere x2=y2⇒x=±y⇒f is not one - one Step 2 To check: f is onto∃y∈R such that f(x)≠y,∀x∈RReason: suppose y=−1∈R such that x2≠−1 (or) x2=−1∈RTherefore f is not onto and hence f is not bijective

sa1yap80 · Accepted Answer

b) Step 3Given: F={y∣y=ax3+b,a,b∈R and x∈R}Claim: To prove that F=RIt si similiar to prove that F is onto(i.e..) ∃y∈R such that F(x)=y∀x∈RHence we can conclude that F=R

a) Consider the following functional relation, f, defined

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