# Functions

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• Sep 29th 2005, 01:20 AM
MMcCann
Functions
Hello,
For some reason these three problems have been giving me trouble. They deal with functions and models. Can anyone help me out with them? If you can, show me all of the steps involved, since that usually helps me see where I am going wrong.

1. Find the domain and range of g(x) = square root of 2 + x + square root of 1 – x
(To clarify, after the equals sign the square root signs on both sides of the addition sign cover the 2+x and 1-x respectively)

2. Find the equation of the line with slope 3, which passes through the point (3,5). Write your answer in slope-intercept form.

3. Let ƒ(x) = x^2 + 3x – 2 Find and simplify (To clarify, after the equal sign it is x to the second power plus 3x minus two)

a. ƒ(x) + h
b. ƒ(x + h)
c. ƒ(2x)
d. ƒ(x + h) - ƒ(x - h)
• Oct 6th 2005, 09:02 AM
TheBrain
Hi, it may be too late but I think I can provide some help on this one.

1/ root(2+x) + root(1-x)
Domain - the numbers that you can set x as. You cannot have the square root of a negative number, so the range will be anything that makes (1-x) ≥ 0 and (2+x) ≥ 0
Answer : -2 ≤ x ≤ 1
Range - the range of numbers that you can get out of the equation. This is suppose would depend if you are allowing -ve roots, you should be able to get an answer from this though.

Note : I occasionally get range + domain backwards so it may be worth double checking.

2/ Passes through (3,5), gradient = 3
Hmm. Try this :

(Y-Y1) / (X-X1) = gradient

In this case, Y1 = 5, Y2 = 3, gradient = 3.

3/ ƒ(x) = x² + 3x – 2
a. ƒ(x) + h - This is just x² + 3x – 2 + h
b. ƒ(x + h) - This means just replace x with (x+h) => (x+h)² + 3(x+h) – 2
c. ƒ(2x) - Just replace x with 2x => (2X)² + 3(2X) – 2
d. ƒ(x + h) - ƒ(x - h) - Similar to part b really =>
[ (x+h)² + 3(x+h) – 2 ] - [ (x-h)² + 3(x-h) – 2 ]

Parts c + d can defaintely be simlified but shouldn't be too difficult once you have muliplied out the brackets.

Mark