exam1_2009_sol

# Solution we complete the square to write the

This preview shows pages 1–4. Sign up to view the full content.

Solution: We complete the square to write the quadratic function f ( x ) in ver- tex form. f ( x ) = - 3 x 2 + 6 x + 2 = - 3[ x 2 - 2 x ] + 2 = - 3[( x - 1) 2 - ( - 1) 2 ] + 2 = - 3[( x - 1) 2 - 1] + 2 = - 3( x - 1) 2 + 3 + 2 = - 3( x - 1) 2 + 5 . So the vertex of the graph of f is (1 , 5)

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

View Full Document
Math 171: Exam 1 (b) [2 points] Does f have a minimum or maximum value? If so, what is it? Solution: Since the coefficient of x 2 is negative, we know that f has a maximum value and no minimum value. The maximum value occurs at the vertex, so it is f (1) = 5. 4. [10 points] Find two numbers whose sum equals 6 and whose product equals 3. Solution: We want two numbers a and b such that: a + b = 6, and ab = 3 . Solving the first equation for b gives us b = 6 - a . Substituting into the second equation, we get a (6 - a ) = 3. Rearranging gives us - a 2 + 6 a - 3 = 0. Using the quadratic formula to solve for a , we have a = - 6 ± p 6 2 - 4( - 1)( - 3) 2( - 1) = - 6 ± 24 - 2 = 3 6 . Taking a = 3 + 6 we get b = 6 - a = 6 - (3 + 6) = 3 - 6. So the two numbers are 3 + 6 and 3 - 6. It is easy to check that these numbers work. 5. Let f ( x ) = x 1 / 3 + 5 2 x 1 / 3 - 1 . (a) [7 points] Find a formula for f - 1 ( x ). Solution: We set y equal to f ( x ) and solve for x : y = x 1 / 3 + 5 2 x 1 / 3 - 1 y (2 x 1 / 3 - 1) = x 1 / 3 + 5 2 yx 1 / 3 - y = x 1 / 3 + 5 2 yx 1 / 3 - x 1 / 3 = y + 5 x 1 / 3 (2 y - 1) = y + 5 x 1 / 3 = y + 5 2 y - 1 x = ( x 1 / 3 ) 3 = y + 5 2 y - 1 3 . Page 2
Math 171: Exam 1 So f - 1 ( x ) = x + 5 2 x - 1 3 .

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 ]}

### 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