Answer:
a.)A and B are the intersection between the two curves. The x coordinate is the equation
[tex]f(x) = g(x)[/tex]
Let's solve it
[tex]\frac12 (x-2)^3 +1 = 2x^2-6x-3\\x^3-6x^2+12x-8+2=4x^2-12x-6\\x^3-10x^2+24x=0\\x(x^2-10x+24)=0\\x(x-4)(x-6)=0[/tex]
We can see that if we substitute x= 0 or x= 4 the equation is verified.
b. The equation has a third solution, x = 6, that is too far to the right to be visible in the graph.
c. the x coordinate of C is the solution of [tex]f(x)=0[/tex] or [tex]\frac 12 (x-2)^3 +1=0\\(x-2)^3 = -2\\(x-2) = -\sqrt[3]2[/tex]
At this point let's guesstimate the cubic root of 2 -let's say the cubic root of 2 is equal to 1plus an error e, and since we know how to calculate the cube of a binomial, let's build it.
[tex](1+e)^3 = 2 \rightarrow 1+3e+3e^2+e^3 = 2[/tex] Now, assuming the error is small enough, it's square and its cube gets even smaller. (if the error is 0.1, it's square becomes 0.01 and the cube 0.01), let's ignore them. [tex]1+3e=2 \rightarrow 3e=1 e= \frac13[/tex] So, going back to our equation, we can tell that the right hand side is [[tex]\mathbb{R}[/tex]]-(1+\frac13)[/tex] Not perfect, but good enough. Let's solve it now.
[tex]x-2= -1-\frac13 \rightarrow x=1-\frac13 = \frac23[/tex]
Which seems close enough from the photo.
c. Domain of both function are the whole Real set, since they're polynomials. Range of f(x) is again the whole real set, since it's an odd degree polynomial. g(x) is a parabula and its range is for y above the vertex y-coord, which is [tex]y>g(\frac32) = 2(\frac94)-6(\frac32)-3 = \frac92 -3(\frac62) -\frac62 = \frac92-4\frac62= -\frac{24-9}2 = -\frac{15}2[/tex]
