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

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Discussion Overview

The discussion revolves around solving the expression (f + g - h)(5) using the functions f(x) = x + 2, g(x) = 5x, and h(x) = 5. The focus is on understanding the evaluation of these functions and the arithmetic involved in combining their outputs.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in solving the problem and states that the result equals 27.
  • Another participant explains that the expression (f + g - h)(5) can be rewritten as f(5) + g(5) - h(5) and provides the step-by-step evaluation leading to 27.
  • A different participant outlines two methods to arrive at the same result: one by evaluating each function at x = 5 and another by simplifying the expression to a general form (f + g - h)(x) = 6x - 3, then substituting x = 5.
  • A later reply acknowledges the help received and shares their experience of self-teaching precalculus, indicating a personal struggle but progress in understanding.

Areas of Agreement / Disagreement

Participants generally agree on the evaluation process and the final result of 27, but there is some ambiguity regarding the notation of F, G, and H, which is not clarified in the discussion.

Contextual Notes

There is a lack of clarity regarding the definitions of F, G, and H, as well as the potential implications of using different methods to solve the problem. The discussion does not resolve these aspects.

Who May Find This Useful

Individuals studying precalculus or seeking assistance with function evaluation and algebraic manipulation may find this discussion beneficial.

sfeld
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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
 
Last edited:
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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:

[tex](f + g - h)(5)[/tex]

is a condensed way of writing:

[tex]f(5) + g(5) - h(5)[/tex]

[tex]= (5+2) + 5(5) - 5[/tex]

[tex]= 7 + 25 - 5[/tex]

[tex]= 27[/tex]
 
Last edited:
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!
 
thanks for the help, sorry, I am teaching my self pre calc, its hard, but I'm getting some where!
 

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