{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Section 4.1-solutions

# Section 4.1-solutions - to(aqt73 Section 4.1 isaacson(55826...

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

to (aqt73) – Section 4.1 – isaacson – (55826) 1 This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Find all the critical points of f when f ( x ) = x x 2 + 25 . 1. x = 0 , 5 2. x = - 25 , 25 3. x = - 5 , 0 4. x = - 5 , 25 5. x = - 25 , 5 6. x = - 5 , 5 correct Explanation: By the Quotient Rule, f ( x ) = ( x 2 + 25) - 2 x 2 ( x 2 + 25) 2 = 25 - x 2 ( x 2 + 25) 2 . Since f is differentiable everywhere, the only critical points occur at the solutions of f ( x ) = 0, i.e. , at the solutions of 25 - x 2 = 0 . Consequently, the only critical points are x = - 5 , 5 . 002 10.0points Find all the critical points of the function f ( x ) = 2 sin x - | x | on the interval ( - π, π ). 1. x = 0 2. x = - π 3 , 0 , 2 π 3 3. x = - 2 π 3 , π 3 4. x = - 2 π 3 , 0 , π 3 correct 5. x = - π 6 , 0 , π 6 6. x = - 5 π 6 , - π 6 , 0 , π 6 , 5 π 6 7. x = - 5 π 6 , 0 , 5 π 6 8. x = - π 3 , 2 π 3 Explanation: Since | x | is differentiable everywhere except at x = 0, while sin x is differentiable for all x , the point x = 0 is a critical point of f and all other critical points will be the solutions of f ( x ) = 0. Now f

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}