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## Math 158 exam 1 study guide

by: Shayla Pedigo

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# Math 158 exam 1 study guide Math 15800

Marketplace > Purdue University > Mathmatics > Math 15800 > Math 158 exam 1 study guide
Shayla Pedigo
Purdue

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These are good notes to study before your first exam.
COURSE
Math
PROF.
David Norris
TYPE
Study Guide
PAGES
4
WORDS
CONCEPTS
Math, Math 158, Exam 1
KARMA
50 ?

## Popular in Mathmatics

This 4 page Study Guide was uploaded by Shayla Pedigo on Thursday February 4, 2016. The Study Guide belongs to Math 15800 at Purdue University taught by David Norris in Spring 2016. Since its upload, it has received 59 views. For similar materials see Math in Mathmatics at Purdue University.

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Date Created: 02/04/16
MATH 158 Exam 1 Study guide Functions inputs Example 1.) f(x)= 2x+3 f(0), f(1), f(a) =2(0)+3 =3 =2(1)+3 =5 outputs =2(a)+3 =2a+3 Example 2.) f(x)=√2x+3 =f(2) = √2(2)+3 4+3 = √ = √7 Domain and Range Example 1.) Let f(x)=1 3x-1 Find domain of f 0 -1 output will be DNE 1 ½ when we divide by 0 1/3 DNE Not in the Domain: domain Because -∞ 1/3 ∞ it is DNE (-∞, 1/3) U (1/3, ∞) Example 2.) f(x) = 1 2 2 x -3x+2 =x -3x+2=0 Find domain of f =(x-2)(x-1)=0 x=2, x=1 Domain: -∞ 1 2 ∞ (-∞,1) U (1,2) U (2, ∞) If f is a fraction, check for domain problems when dividing by 0 (i.e. when denominator is 0) Example 3.) f(x) = √ x fraction part: x2 divide by 0 when denominator is 0 if and only if x2 – 1 = 0 (x+1) (x- 1)=0 x=-1, x=1 ∞ -1 0 1 ∞ √ part: Domain: [0,1) U (1, ∞) take square root of negative number when square root has a negative number if and only if x < 0. Application of functions Example 1.) A ball is dropped and its height t seconds after dropping is modeled by 2 f(t) = -t +11 a.) Find Domain: [0,∞) >>Because t means time after ball was dropped, t can’t be negative! b.) What is the height of the ball 1 second after dropping? f(1) = -12 +11 =-1+11 =10 height c.) When is the ball at height 2? f(t) = 2 - t2 +11 = 2 t2 11=2+ 2 0= t -9 t can’t 0=(t+3) 0=(t-3) be negativ t = -3 t = 3 e t2 Example 3.) A farmer fences in a rectangular area of area 10with 5 foot gates as shown: 1.) Write l in terms of w A = (l) (w) 100 = (l) (w) w w 100 = l w 2.) How much fencing is used? = l + l + (w-g) + (w-g) + (w-g) (write as function of w) = 2l + 3(w-g) = 2l+3(w-5) Replace l with 100 from # 1. = 2(100) + 3(w-5) Put 5 in because 5ft = gates w 2w 3.) If the fencing is \$10 a foot, what is the total cost of fencing? = (10) [ 2 (100) + 3(w-5)] w Simplifying Algebraic Expressions Example 1.) 3 – 1 1. Simplify Numerator 5 10 2 3 - 1 1 + 1 = 2 5 10 2 3 1 + 1 2 3 = 6 - 1 10 10 1 + 1 2 3 5 = 10 1 + 1 2 3 2. Simplify Denominator 5 = 10 3 1 + 1 1 3 2 3 3 5 = 10 3 + 2 6 6 5 = 10 2 6 3. Multiply by reciprocal 5 x 6 = 6 = 3 10 5 10 5 Function Arithmetic and Composition of Functions Example 1.) f(x)= 2x+3; g(x)= x+4 Find f(g(1)= f(5)= 13 Plug in 5 Find (g o f)(1) = g(f(1))= g(2(1)+3) g(5) 5+4 = 9 Write (f o g)(x) = f(g(x)) = f(x+4) = 2(x=4)+3 =2x+8+3 = 2x+11

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