practice test 2 answers

practice test 2 answers - MAT 111 Practice Test 2 Solutions...

This preview shows pages 1–5. Sign up to view the full content.

MAT 111 Practice Test 2 Solutions Spring 2010 1

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

View Full Document
1. (10 points) Find the equation of the tangent line to 2 x + 2 y = 1+ x 2 y 2 at the point (1 , 1). The equation is y - y 0 = dy dx ( x - x 0 ) So all we need is dy/dx . Differentiating the above equation implicitly, we get 1 2 2 x + 2 y (2 + 2 dy dx ) = 2 xy 2 + 2 y dy dx x 2 So, substituting x = 1 and y = 1, we get 1 4 (2 + 2 dy dx ) = 2 + 2 dy dx solving, we get dy dx = - 1. So the equation of the tangent is y - 1 = - 1( x - 1) OR y = - x + 2 2
2. (30 points) Evaluate the following: (a) lim x 2 x 4 - 2 x - 12 x - 2 Solution: Notice that if we let f ( x ) = x 4 - 2 x then f (2) = 12 and so the above limit is actually equal to lim x 2 f ( x ) - f (2) x - 2 which is precisely f (2). Thus, all we need to do is evaluate f ( x ) and substitute 2. f ( x ) = 4 x 3 - 2 and so lim x 2 f ( x ) - f (2) x - 2 = f (2) = 4 · 2 3 - 2 = 30 . (For the PE class students, the same solution with f ( x ) = x 4 + 2 x and f ( x ) = 4 x 3 + 2 and limit = 34). (b) Find f ( x ) if f ( x ) = sec 3 x 1+sin x Solution: By Quotient Rule, f ( x ) = (1 + sin x ) d dx (sec 3 x ) - sec 3 x d dx (1 + sin x ) (1 + sin x ) 2 By Chain Rule d dx (sec (3 x )) = sec (3 x ) tan (3 x ) d dx (3 x ) = 3 sec (3 x ) tan (3 x ). So the final answer is: 3 sec (3 x ) tan (3 x )(1 + sin x ) - cos ( x ) sec (3 x ) ( i + sin x ) 2 (c) Find f ( x ) if f ( x ) = sin 2 x x 3 +1 cos x +2 . Solution: This is just a repeated application of chain rule, quotient rule and product rule: Outer most functions is square, next is sin and then there is a quotient to be differentiated. f ( x ) = 2 sin x x 3 + 1 cos x + 2 d dx sin x x 3 + 1 cos x + 2 3

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

View Full Document
= 2 sin x x 3 + 1 cos x + 2 cos x x 3 + 1 cos x + 2 d dx x x 3 + 1 cos x + 2 = 2 sin x x 3 + 1 cos x + 2 cos x x 3 + 1 cos x + 2 (cos x + 2) d dx ( x x 3 + 2) - x x 3 + 2 d dx (cos x + 2) (cos x + 2) 2 = 2 sin x x 3 + 1 cos x + 2 cos x x 3 + 1 cos x + 2 · (cos x + 2)( x 3 + 1 + x 1 2 x 3 +1 (3 x 2 )) - x x 3 + 2( - sin x ) (cos x + 2) 2 = 2 sin x x 3 + 1 cos x + 2 cos x x 3 + 1 cos x + 2 (cos x + 2)(
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