For the following functions from R -> R, determine if function is one to one, onto, or both. Explain.a) f(x)=3x-4b)g(x)=(x^2)-2c) h(x)=2/xd) k(x)=ln(x)e) l(x) = e^x

Answer with explanation:a. f(x)=3x-4Let [tex]f(x_1)=f(x_2)[/tex][tex]3x_1-4=3x_2-4[/tex][tex]3x_1=3x_2-4+4[/tex][tex]3x_1=3x_2[/tex][tex]x_1=x_2[/tex]Hence, the function one-one.Let f(x)=y  [tex]y=3x-4[/tex][tex]3x=y+4[/tex][tex]x=\frac{y+4}{3}[/tex]We can find pre image in domain R for every y in range R.Hence, the function onto.b.g(x)=[tex]x^2-2[/tex]Substiute x=1Then [tex]g(x)=1-2=-1[/tex]Substitute x=-1Then g(x)=1-2=-1Hence, the image of 1 and -1 are same . Therefore, the given function g(x) is not one-one.The given function g(x) is not onto because there is no pre image of -2, -3,-4......  R.Hence, the function neither one-one nor onto on given  R.c.[tex]h(x)=\frac{2}{x}[/tex]The function is not defined for x=0 .Therefore , it is not a function on domain R.Let [tex]h(x_1)=h(x_2)[/tex][tex] \frac{2}{x_1}=\frac{2}{x_2}[/tex]By cross mulitiply [tex]x_1= \frac{2\times x_2}{2}[/tex][tex]x_1=x_2[/tex]Hence, h(x) is a one-one function on R-{0}.We can find pre image for every value of y except zero .Hence, the functionh(x) is onto on R-{0}.Therefore, the given function h(x) is both one- one and onto on R-{0} but not on R. d.k(x)= ln(x)We know that logarithmic function not defined for negative values of x. Therefore, logarithmic is not a function R.Hence, the given function K(x) is not a function on R.But it is define for positive R.Let[tex]k(x_1)=k(x_2)[/tex] [tex] ln(x_1)=ln(x_2)[/tex]Cancel both side log then [tex]x_1=x_2[/tex]Hence, the given function one- one on positive R.We can find pre image in positive R for every value of [tex]y\in R^+[/tex].Therefore, the function k(x) is one-one and onto on [tex]R^+[/tex] but not on R.e.l(x)=[tex]e^x[/tex]Using horizontal line test if we draw a line y=-1 then it does not cut the graph at any point .If the horizontal line cut the graph atmost one point the function is one-one.Hence, the horizontal line does not cut the graph at any point .Therefore, the function is one-one on R.If a horizontal line cut the graph atleast one point then the function is onto on a given domain and codomain.If we draw a horizontal line y=-1 then it does not cut the graph at any point .Therefore, the given function is not onto on R.