Homework 9-solutions

Homework 9-solutions - wtm369 – Homework 9 – Helleloid...

This preview shows pages 1–3. Sign up to view the full content.

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.

Unformatted text preview: wtm369 – Homework 9 – Helleloid – (58250) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The derivative of a function f is given for all x by f ′ ( x ) = (3 x 2 + 3 x − 36) parenleftBig 1 + g ( x ) 2 parenrightBig where g is some unspecified function. At which point(s) will f have a local maximum? 1. local maximum at x = − 3 2. local maximum at x = − 4 correct 3. local maximum at x = − 4 , 3 4. local maximum at x = 4 5. local maximum at x = 3 Explanation: At a local maximum of f , the derivative f ′ ( x ) will be zero, i.e. , 3( x − 3)( x + 4) parenleftBig 1 + g ( x ) 2 parenrightBig = 0 . Thus the critical points of f occur only at x = − 4 , 3. To classify these critical points we use the First Derivative test; this means looking at the sign of f ′ ( x ). But we know that 1 + g ( x ) 2 is positive for all x , so we have only to look at the sign of the product 3( x − 3)( x + 4) of the other two factors in f ′ ( x ). Now the sign chart − 4 3 + + − for 3( x − 3)( x + 4) shows that the graph of f is increasing on ( −∞ , − 4), decreasing on ( − 4 , 3), and increasing on (3 , ∞ ). Thus f has a local maximum at x = − 4 . 002 10.0 points Let f be the function defined by f ( x ) = 5 − x 2 / 3 . Consider the following properties: A. concave down on ( −∞ , 0) ∪ (0 , ∞ ) ; B. derivative exists for all x negationslash = 0 ; C. has local maximum at x = 0 ; Which does f have? 1. All of them 2. B only 3. None of them 4. B and C only correct 5. C only 6. A and B only 7. A and C only 8. A only Explanation: The graph of f is 2 4 − 2 − 4 2 4 On the other hand, after differentiation, f ′ ( x ) = − 2 3 x 1 / 3 , f ′′ ( x ) = 2 9 x 4 / 3 . Consequently, A. not have: ( f ′′ ( x ) > , x negationslash = 0); wtm369 – Homework 9 – Helleloid – (58250) 2 B. has: ( f ′ ( x ) = − (2 / 3) x − 1 / 3 , x negationslash = 0); C. has: (see graph). 003 10.0 points Use the graph a b c of the derivative of f to locate the critical points x at which f does not have a local maximum? 1. x = b , c 2. x = c 3. x = b 4. none of a , b , c 5. x = a 6. x = c , a correct 7. x = a , b 8. x = a , b , c Explanation: Since the graph of f ′ ( x ) has no ‘holes’, the only critical points of f occur at the x- intercepts of the graph of f ′ , i.e. , at x = a, b, and c . Now by the first derivative test, f will have (i) a local maximum at x if f ′ ( x ) changes from positive to negative as x passes through x ; (ii) a local minimum at x if f ′ ( x ) changes from negative to positive as x passes through x ; (iii) neither a local maximum nor a local minimum at x if f ′ ( x ) does not change sign as x passes through x ....
View Full Document

{[ snackBarMessage ]}

Page1 / 8

Homework 9-solutions - wtm369 – Homework 9 – Helleloid...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online