Having A really hard time and my Professor is being a A Hole with help so can someone just solve it for me I need the range and the graph.​

Having A really hard time and my Professor is being a A Hole with help so can someone just solve it for me I need the range and the graph class=

Respuesta :

Normally, when [tex]x=3[/tex] the function

[tex]f(x)=\dfrac12(x-1)^3-1[/tex]

takes on a value of

[tex]f(3)=\dfrac12(3-1)^3-1=\dfrac82-1=4-1=3[/tex]

so the graph of [tex]f(x)[/tex] passes through the point (3, 3). This is the only place at which the cubic takes on a value of 3, since [tex]f(x)[/tex] is one-to-one.

In the definition of [tex]q(x)[/tex], however, we remove this point from the curve and replace with the point (3, -2).

While [tex]f(x)[/tex] is a standard cubic function, so that its range is all real numbers, the replacement made for [tex]q(x)[/tex] removes 3 from the range set, so that the range of [tex]q(x)[/tex] would be all real numbers *except* 3.