{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW5 Solutions

# HW5 Solutions - ASSIGNMENT 5 SOLUTIONS Problem 1 Done in...

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

ASSIGNMENT 5 - SOLUTIONS Problem 1. Done in class. Problem 2. Take f ( x ) = x 2 , whose graph is a parabola. Drop it by 1 (so now the function is g ( x ) = x 2 - 1). Slide it to the right by 1 (the function becomes h ( x ) = ( x - 1) 2 - 1 = x 2 - 2 x .) This graph has all the required properties. Problem 3. By definition, concave upwards means increasing first derivative. If f and g are increasing, then so is their sum f + g = ( f + g ) . So the first statement is true. But the second is false. For example, both f ( x ) = x 2 and g ( x ) = x 4 + x 2 are concave upwards on [0 , 1] yet f ( x ) - g ( x ) = - x 4 is concave downwards on [0 , 1]. Problem 4. To start off, the domain of f ( x ) = x 9 - x is ( -∞ , 9]. The first derivative is f ( x ) = x 9 - x + x ( 9 - x ) = 9 - x + x 1 2 1 9 - x (9 - x ) = 9 - x - x 2 9 - x = 2(9 - x ) - x 2 9 - x = 18 - 3 x 2 9 - x = 3 2 6 - x 9 - x The critical numbers of f ( x ) are x = 6 ( f (6) = 0) and x = 9 ( f ( x ) DNE). Both points are in the domain of f ( x ). i) Let’s use the 1st derivative test. At x = 6, f ( x ) changes sign from + to - , so f (6) = 6 3 is a relative maximum. At x = 9 there is no change in the sign of f ( x ) because we cannot go past 9. So, strictly speaking, the 1st derivative test does not apply at this point. However, we
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online