sol12-2 - and so y = 7 2 x 5 2 1 2 x 1 2 = = 7 2 x 5 2 1 2...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
SOLUTIONS CHAPTER 12.2 MATH 132 WI01 2. Proof. f ( x ) is a constant! it does not contain any x in its formula, hence f 0 ( x ) = 0 8. Proof. y 0 = (4 · x 4 ) 0 = 4( x 4 ) 0 = 4 · 4 x 3 = 16 x 3 28. Proof. f ( x ) = 5( x 4 - 3) 2 = 5 2 ( x 4 - 3) and so f 0 ( x ) = ( 5 2 ( x 4 - 3)) 0 = 5 2 ( x 4 - 3) 0 = 5 2 (4 x 3 ) = 10 x 3 48. Proof. y = 1 4 x 5 = 1 4 · 1 x 5 = 1 4 x - 5 and so y 0 = ( 1 4 x - 5 ) 0 = 1 4 ( x - 5 ) 0 = 1 4 ( - 5) x - 5 - 1 = - 5 4 x - 6 74. Proof. f ( x ) = 7 x 3 + x 2 x = 7 x 3 2 x + x 2 x = = 7 2 · x 3 x 1 2 + 1 2 · 1 x 1 2 = = 7 2 x 3 - 1 2 + 1 2 x 1 - 1 2 = 7 2 x 5 2 + 1 2 x 1 2 Date : 01/21/2000. 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 MATH 132 WI01
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: and so y = ( 7 2 x 5 2 + 1 2 x 1 2 ) = = ( 7 2 x 5 2 ) + ( 1 2 x 1 2 ) = = 7 2 · ( x 5 2 ) + 1 2 · ( x 1 2 ) = = 7 2 · 5 2 x 5 2-1 + 1 2 · 1 2 x 1 2-1 = = 35 4 x 3 2 + 1 4 x-1 2 ±...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern