Given g(x) = (x + 1)(x^2 − x), g: R → R where R is the set of real numbers.
a) Find the domain and range of the function g. (2 marks)
b) Determine whether the function is injective, surjective, and/or bijective. Justify
your answers. (7 marks)
a:Dom(g)=R,since g is defined for all xlimx→+∞(x+1)(x2−x)=[+∞⋅+∞]=+∞limx→−∞(x+1)(x2−x)=[+∞⋅−∞]=−∞Since g is continuous,from these Range(g)=Rb:injective − false:g(0)=g(1)=0surjective − true,since Range(g)=Rbijective − false,since it is not injectivea:\\Dom\left( g \right) =\mathbb{R} , \sin ce\,\,g\,\,is\,\,defined\,\,for\,\,all\,\,x\\\underset{x\rightarrow +\infty}{\lim}\left( x+1 \right) \left( x^2-x \right) =\left[ +\infty \cdot +\infty \right] =+\infty \\\underset{x\rightarrow -\infty}{\lim}\left( x+1 \right) \left( x^2-x \right) =\left[ +\infty \cdot -\infty \right] =-\infty \\Since\,\,g\,\,is\,\,continuous, from\,\,these\,\,Range\left( g \right) =\mathbb{R} \\b:\\injective\,\,-\,\,false: g\left( 0 \right) =g\left( 1 \right) =0\\surjective\,\,-\,\,true, \sin ce\,\,Range\left( g \right) =\mathbb{R} \\bijective\,\,-\,\,false, \sin ce\,\,it\,\,is\,\,not\,\,injectivea:Dom(g)=R,sincegisdefinedforallxx→+∞lim(x+1)(x2−x)=[+∞⋅+∞]=+∞x→−∞lim(x+1)(x2−x)=[+∞⋅−∞]=−∞Sincegiscontinuous,fromtheseRange(g)=Rb:injective−false:g(0)=g(1)=0surjective−true,sinceRange(g)=Rbijective−false,sinceitisnotinjective
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