Show What You Know: Manipulating Expressions from Calculus MAT 190 Precalculus Objectives: The purpose of this assignmen
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Show What You Know: Manipulating Expressions from Calculus MAT 190 Precalculus Objectives: The purpose of this assignment is for you to: 1. demonstrate your ability to algebraically manipulate expressions from calculus; 2. improve your mathematical writing to include full solutions with justifications; 3. integrate mathematical statements into grammatically correct expositions. Assignment: Algebraically manipulate ONE of the following expressions. Include a detailed explanation for each mathematical step written in grammatically correct complete sentences within a 2-column format. Your final answer should follow these guidelines: Completely factored • Reduced into lowest terms • Single quotient . Only positive exponents and roots 2(2 – 2c+5)(2r – 2)2 – (r2 – 2c+5),2(2) 1 - 1. elx 2. 3(2x + 1)² (2) cos(r) + (2x+1)³(4) cos³(x)(-sin(x)) 4(6x − 5)³(6) (x² + 1) — (6x - 5)¹ (²) (x² + 1)² (2x) 3. (7²+1)3
Solution: 3e³ (6²+1)+() (6x²+1)-(12) 3³ (6r²+1)+6³ (62²+1)- 62e³ 3e (62²+1) +- (6x² + 1) 3³ (6x²+1). (6² +1}} + Gre (6x2+1) (6r²+1) 62e³ 3 (6r2+1)(6r²+1) (6x² + 1) (6x²+1) 3e³ (6²+1) (6x² + 1) + + 6re (6x² +1) This is the expression we were given to manipulate. In the second term of the sum, we multiplied the factor and the factor 12r and placed the re- sult at the front of the second term using the commutative and associative properties of multi- plication. In the second term of the sum, we rewrote the factor (62²+1)- using a property of as (6x²+1) exponents. To get a common denominator for the two terms of the sum, we multiplied the first term in the (²+1) sum by a special form of 1. We completed the multiplication of fractions in the first term of the sum. We completed the multiplication of the two factors with the same base in the numerator of the first term of the sum by adding the exponents.
3e³ (6x2 + 1) +62e³ (6x² + 1) 3e ((6r2+1)+2x) (6x² + 1) 3e³ (6r²+2r+1) (6x² +1) We added the two fractions with a common denominator by adding the numerators and plac- ing the sum on top of the com- mon denominator. We factored out the greatest common factor from the two terms in the sum in the numer- ator. We rearranged the terms in the second factor in the numerator using the commutative property. of addition. Since the second factor in the numerator cannot. be factored and the fraction can- not be reduced, we are done.