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# MATH121, Lesson 4.1 Notes Math 121

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This 3 page Class Notes was uploaded by Mallory McClurg on Friday September 23, 2016. The Class Notes belongs to Math 121 at University of Mississippi taught by Dirle in Fall 2016. Since its upload, it has received 30 views. For similar materials see College Algebra in Math at University of Mississippi.

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Date Created: 09/23/16

MATH121 Chhaapptteerr 4 NNOOT TEES S Lesson 4.1 – Relations and Functions EXAMPLE 1. f(x) = √(x – 1) + 6 (We’re asked to find the “implied domain” of this function. Basically that means…what all can x equal to keep the function “defined”? Remember, with a square roo t symbol, the only way to keep the function defined is to make sure that what’s under the square root symbol is positive. Which numbers can you subtract 1 from and still get a positive number? Write it in interval notati on.) [ 1, ∞) (All numbers, including 1, from 1 to infinite will keep the function defined.) EXAMPLE 2. 2 g(x) = -5x – 3x (Look at all of the parts of the problem with an x. Is there any reason that this problem would be undefined? Remember, with exponents that are multiples of 2 , that means your answer will always be positive; for example…(2) and (-2) are both positive.) (-∞, ∞) or ALL REAL NUMBERS (All real numbers keep the function defined.) EXAMPLE 3. f(x) = -3/(x + 8) (Look at the parts of the problem with an x. Here, there is an x in the denominator of a fraction. We know that if the denominator of a fraction is 0, then it’s undefined. So what values of x would keep the deno minator NOT equal to zero?) x + 8 ≠ 0 x ≠ -8 (These computations help us determine that our interval notation will NOT include -8.) (-∞, -8) U (-8, ∞) (This means that x = everything except -8.) EXAMPLE 4. f(x) = (√(x – 7))/-8 (Look at the parts of the problem with an x. We want the expression in the numerator of the fraction to be greater than or equal to zero.) x – 7 ≥ 0 x ≥ 7 (These are the computations we need to perform in order to find out what our interval notation will include.) [7, ∞) (The function is defined everywhere from 7 inclusively, to positive infinite.) MATH121 Chhaapptterr 4 NOOT TEES S EXAMPLE 5. f(x) = √(x + 5) – 6 (The question here asks us to find the function of f(x – 1), which means all we have to do is substitute “x – 1” everywhere we see an x in the problem.) f(x – 1) = √((x – 1) + 5) – 6 (Combine the like terms under the square root symbol to simplify.) f(x – 1) = √(x + 4) – 6 (This is our answer!) EXAMPLE 6. g(x) = x + 3 (The question asks us to find the function of 2 2 g(x ), so substitute an x everywhere you see an x in the function.) g(x ) = (x ) + 3 (Simplify.) 2 4 g(x ) = x + 3 (This is our answer!) EXAMPLE 7. 2 h(x) = -3x / (x – 2x + 1) (The question asks us to find the function of h(x – 1), so substitute “x – 1” everywhere there’s an x.) h(x – 1) = -3(x – 1) / (x – 1) – 2(x – 1) + 1 (To simplify, we can distribute the -3 to the x and -1 in the numerator. Then, FOIL out the (x – 1) in the denominator, and also distribute the -2 to the x and -1 in the denominator. Combine like terms to simplify even more.) h(x – 1) = (-3x + 3) / (x – 4x + 4) (This is our simplified answer!) EXAMPLE 8. f(x) = √(x – 6) + 1 (The question asks us to find the simplified function of “f(x + a) – f(x)”, so in the fist part of our answer, we need to substitute (x + a) where there is an x, and then subtract that function with the original f(x) function.) f(x + a) – f(x) = √(x + a – 6) + 1 – (√(x – 6) + 1) (Simplify. Don’t forget, you’ll need to change the signs in the second part of the problem, because we’re subtracting the entire expression.) f(x + a) – f(x) = √(x + a – 6) – √(x – 6) (This is our simplified answer!) MATH121 Chhaapptterr 4 NO OTTEESS EXAMPLE 9. g(x) = x + 3 (The question asks us to find he function of “g(x + a) – g(x)”, so do what you did in the last problem and substitute “x + a” in for x, then subtract the original function.) g(x + a) – g(x) = x + a + 3 – x – 3 (Combine like terms.) g(x + a) – g(x) = a (This is our simplified answer!) EXAMPLE 10. g(x) = 2x –x + 4 (The question asks us to find the function of “g(x + a) – g(x)”, so we’re doing the same thing, yet again here.) 2 2 2 g(x + a) – g(x) = 2(x + a) – (x + a) + 4 – (2x – x + 4) (FOIL out the (x + a) and then distribute the 2 to all the FOILed values.) g(x + a) – g(x) = 2x + 4ax + 2a – x – a + 4 – 2x + x – 4 2 (Combine like terms.) 2 g(x + a) – g(x) = 4ax + 2a – a (This is our simplified answer!) EXAMPLE 11. 2 f(x) = 3x + 3x – 4 (The question asks us to find the function of “(f(x + h) – f(x))/ h”, so using substitution, find our new function.) f(x + h) – f(x) / h = (3(x + h) + 3(x + h) – 4 – (3x + 3x – 4)) / h) 2 (FOIL out the (x + h) and then distribute the 3 to all of the numbers. Then distribute the 3 to the x and the h. Don’t forget, when subtracting the original function at the end of the numerator, change the signs of each of the numbers.) f(x + h) – f(x) / h = (3x + 6xh + 3h + 3x + 3h – 4 – 3x – 3x + 4) / h 2 (Combine like terms in the numerator.) 2 f(x + h) – f(x) / h = (6xh + 3h + 3h) / h (Now, take one h out of everything in the numerator, to simplify the fraction.) f(x + h) – f(x) / h = 6x + 3h + 3 (This is our simplified answer!)

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