Solve (f+g-h)(5): Precalc Problem

[sfeld]

SOME HOW I tryed a thousand times, the problem equals 27

f(x) = x+2, g(x) = 5x, h(x) = 5
Find (F + G - H)(5)

!PLZ HELP and explain <3

 

[cepheid]

All you are doing is summing up three functions of x. Remember that each of those functions takes a variable x defined on a certain domain and "transforms" it into another variable equal to f(x), which depends on x in a certain way. Since all three functions are defined for all real numbers x, the set of all real numbers is the domain of all three functions. Each function will take on a certain real value for a given real number x (in this case 5) in the domain. You are asked to add up two of these values and subtract from them a third. In other words:

$$(f + g - h)(5)$$

is a condensed way of writing:

$$f(5) + g(5) - h(5)$$

$$= (5+2) + 5(5) - 5$$

$$= 7 + 25 - 5$$

$$= 27$$

 

[HallsofIvy]

There are two ways you can do the problem (seeing that is probably the whole point of assigning the problem!).

First, since f(x)= x+2, f(5)= 5+ 2= 7.
Since g(x)= 5x, g(5)= 25.
Since h(x)= 5 (for all x), g(5= 5.
(f+ g- h)(5)= 7+ 25- 5= 27.
(by the way, after defining f, g, and h, we have no idea what F, G, and H are!)

A more sophisticated way (the way Cepheid did it) is to note that, for any x,
(f+ g- h)(x)= (x+ 2)+ (5x)- 5= 6x- 3 so (f+ g- h)(5)= 6(5)- 3= 30- 3= 27.
By golly both give the same answer!

 

[sfeld]

thanks for the help, sorry, I am teaching my self pre calc, its hard, but I'm getting some where!