MATH
4.4 Review Questions

# 4.4 Review Questions - Solutions to Review Problems for the...

• Notes
• 6

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

Solutions to Review Problems for the Final Exam Solutions to Maxima, minima, convexity, concavity 1. f ( x ) = 2 3 x 3 - x 2 - 4 x + 1, f ( x ) = 2 x 2 - 2 x - 4, f ( x ) = 4 x - 2 (a) f ( x ) = 0 when 0 = x 2 - x - 2 = ( x + 1)( x - 2), i.e., x = - 1, x = 2. (b) f ( - 1) = - 6 < 0, maximum; f (2) = 6 > 0 minimum. (c) increasing for x < - 1, x > 2, (d) convex for x > 1 / 2. 2. f ( x ) = 2 x + 18 /x , f ( x ) = 2 - 18 /x 2 , f ( x ) = 36 /x 3 (a) f ( x ) = 0 when 2 = 18 /x 2 , i.e., x = 3, x = - 3 (b) f ( - 3) = - 36 / 27 < 0, maximum; f (3) = 36 / 27 > 0, minimum (c) increasing for x < - 3, x > 3; (d) convex for x > 0 3. f ( x ) = x x 2 +1 , f ( x ) = ( x 2 +1) - x · 2 x ( x 2 +1) 2 = 1 - x 2 ( x 2 +1) 2 (a) f ( x ) = 0 when x 2 = 1, i.e., x = - 1, x = 1; (c) increasing for - 1 < x < 1; (b) f ( x ) = - 2 x ( x 2 +1) 2 - (1 - x 2 ) · 2( x 2 +1)(2 x ) ( x 2 +1) 4 f ( - 1) = 2 / 2 2 > 0, minimum; f (1) = - 2 / 2 2 < 0, maximum (d) f ( x ) = - 2 x ( x 2 +1)(3 - x 2 ) ( x 2 +1) 4 , convex for - 3 < x < 0, x > 3 4. f ( x ) = - 2 x - 1 / 2 - 3 ln x + 4 x , f ( x ) = x - 3 / 2 - 3 /x + 2 x - 1 / 2 , (a) f ( x ) = x - 1 / 2 ( x - 1 - 3 x - 1 / 2 + 2) = 0 when x - 1 / 2 = 1 , 2 or x = 1 , x = 1 / 4. (b) f ( x ) = - (3 / 2) x - 5 / 2 + 3 x - 2 - x - 3 / 2 , f (1) = - (3 / 2) + 3 - 1 > 0 minimum f (1 / 4) = - (3 / 2)(32) + 3 · 16 - 8 = - 8 < 0 maximum (c) increasing for x < 1 / 4 and x > 1; (d) convex when x - 3 x 1 / 2 + (3 / 2) < 0, that is when r 1 < x < r 2 where r i = (3 ± 9 - 6) / 2 5. f ( x ) = x ( x - 1) 4 , f ( x ) = ( x - 1) 4 + 4 x ( x - 1) 3 = (5 x - 1)( x - 1) 3 f ( x ) = 0 when x = 1 / 5, x = 1. f is increasing for x < 1 / 5, decreasing on (1 / 5 , 1), and increasing for x > 1 so 1/5 is a maximum and 1 is a minimum. 6. (a) h ( t ) = v ( t ) = 10 - 10 t and h (0) = 15, so h ( t ) = 15 + 10 t - 5 t 2 . (b) Highest point is reached when 0 = v ( t ) = 10 - 10 t , i.e., time t = 1. Maximum height h (1) = 15 + 10 - 5 = 20 (c) 0 = 15 + 10 t - 5 t 2 = - 5( t 2 - 2 t - 3 t ) = - 5( t + 1)( t - 3), so t = 3.

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

This is the end of the preview. Sign up to access the rest of the document.
• Spring '07
• DURRETT
• Calculus, Natural logarithm, dx

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern