I am having trouble trying to come up with an introduction and a conclusion for composition and inverse..

Question

I have done the problems but having a hard time with the introduction and conclusion for composition and inverse. Please, please, help

 

 

Answer 1

(f-h)(4)=f(4)-h(4)=2×4+5-(7-4/3)=8+5-7+4/3=6+4/3=7 1/3

 

A.   fοg(x)=f(g(x))=2g(x)+5=2(x²-3)+5=2x²-6+5=2x²-1

B.   hοg(x)=h(g(x))=7-g(x)/3=7-(x²-3)/3=7-x²/3+1=8-x²/3

 

The graph y= g(x) is that of y=x² moved -3 units parallel to the y-axis, so that its vertex is at (0,-3)

The function y=g(x-6)-7 is what is required for the second graph, or more symmetrically written as

y+7=g(x-6)

 

For f(x)=y = 2x+5  we find f^-1(x) n two steps

Solve for x: x=(y-5)/2

Change x to y and y to x   f^-1(x)=y=(x-5)/2

For h(x) = 7-x / 3the procedure is

y-7=-x/3,  -3y+21=x;  y=-3y+21

 

A function is defined by specifying a domain, a co-domain, and a rule for taking a point in the domain and producing a point in the co-domain.  The set of points actually produced is called the range.

The composition of two functions, written for the functions f and g, is the function, NAMED fοg, which is defined, for a point x in the domain of g, as f(g(x)).  It is calculated by computing g(x), and using that value to calculate f(g(x)).  This requires that the domain of f must contain the range of g, else you can't calculate the f.

The inverse of a function takes a value in the range of a function and produces that value in the domain from which it came.  In order to define an inverse, the function itself must be one-one.  The idea is that if we have the function y=x² with domain the real line, and try to invert it, we cannot tell whether the value 16 came from squaring 4 or -4.  The way round this is to restrict the domain, to for example the positive reals.  If we do this then y=x² produces 16 only by squaring 4 and so we may define an inverse b saying that the inverse image of 16 is 4.  In this case f(x)=x², with domain x>=0, has the inverse f^-1(x)=√x, domain x>=0.