MA113_S13_Assignment_1_Solutions

# MA113_S13_Assignment_1_Solutions - Question A Let f(x be...

• Homework Help
• 3

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

Question A: Let f ( x ) be the square of the distance from the point (2 , 1) to a point ( x, 3 x + 2) on the line y = 3 x + 2. Find the minimum value of f ( x ). Solution. The square of the distance from ( x 1 ,y 1 ) to ( x 2 ,y 2 ) is given by d 2 = ( x 2 x 1 ) 2 + ( y 2 y 1 ) 2 , as can be found on page 4 of the textbook. Thus if f ( x ) is defined to be the square of the distance from the point (2 , 1) to the point ( x, 3 x + 2), we have that f ( x ) = ( x 2) 2 + ((3 x + 2) 1) 2 = x 2 4 x + 4 + 9 x 2 + 6 x + 1 = 10 x 2 + 2 x + 5 . We can rewrite this expression by completing the square which can be found on page 18 of the textbook. This yields that f ( x ) = 10 x 2 + 2 x + 5 = 10 parenleftbigg x 2 + 1 5 x parenrightbigg + 5 = 10 parenleftbigg x 2 + 1 5 x + 1 100 1 100 parenrightbigg + 5 = 10 parenleftBigg parenleftbigg x + 1 10 parenrightbigg 2 1 100 parenrightBigg + 5 = 10 parenleftbigg x + 1 10 parenrightbigg 2 + 49 10 . Now we notice that for every x , 10( x + 1 10 ) 2 0. Therefore f ( x ) 49 10 for every x . Since f ( 1 10 ) = 49 10 , we see that the minimum value of f ( x ) in fact is 49 10 . Question B:

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.
• Fall '09
• Calculus, lim, Decimal, Natural logarithm, 8 G, 8 K

{[ 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