{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Exam2_2 3 - Ferentiable everywhere so the only critical...

This preview shows page 1. Sign up to view the full content.

Taylor, Douglas – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: JEGilbert 3 Consequently, dy dx = - 4 x 6 y = - 2 x 3 y . keywords: implicit differentiation 006 (part 1 of 1) 10 points Find the slope of the tangent line to the graph of 2 x 3 + y 3 + xy = 0 at the point P (1 , - 1). 1. slope = - 2 3 2. slope = 4 5 3. slope = 5 4 4. slope = 3 2 5. slope = - 5 4 correct 6. slope = - 4 5 Explanation: Differentiating implicitly with respect to x we see that 6 x 2 + 3 y 2 dy dx + y + x dy dx = 0 . Consequently, dy dx = - 6 x 2 + y 3 y 2 + x . Hence at P (1 , - 1) slope = dy dx fl fl fl P = - 5 4 . keywords: implicit differentiation, tangent line, slope 007 (part 1 of 1) 10 points Locate all the critical points of f ( x ) = ( x + 3) 4 (5 - x ) 3 . 1. x = 5 , - 3 , - 11 7 2. x = 5 , - 3 , 11 7 correct 3. x = 5 , 3 , 11 7 4. x = - 5 , 3 , - 11 7 5. x = 5 , 3 , - 11 7 6. x = - 5 , - 3 , 11 7 Explanation: Since
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ferentiable everywhere, so the only critical points oF f occur at the solutions oF f ( x ) = 4( x + 3) 3 (5-x ) 3-3( x + 3) 4 (5-x ) 2 = 0 . But aFter simpli²cation this becomes ( x + 3) 3 (5-x ) 2 (20-4 x-3 x-9) = ( x + 3) 3 (5-x ) 2 (11-7 x ) = 0 . Consequently, the critical points oF f occur at x = 5 ,-3 , 11 7 . keywords: critical points, polynomial Func-tion, Product Rule 008 (part 1 oF 1) 10 points ±ind the absolute minimum value oF f ( x ) = 1 3 x 3-5 x 2 + 16 x + 4 on the interval [0 , 3]....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online