# Let f x x 3 x 2 2 x on 11 observe that f 1 2 f 1 0

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

Let f ( x ) = x 3 + x 2 - 2 x on [-1,1]. Observe that f (-1) = 2, f (1) = 0, and k = 1 is a number between f (-1) and f (1). Since f is a polynomial, f is continuous on the interval [-1,1]. Consequently, the function f satisfies the hypotheses of the Intermediate Value Theorem. Thus, we are entitled to invoke the magical conclusion that asserts that there is at least one number x 0 in (-1,1) where f( x 0 ) = 1. ______________________________________________________________________ 3. (5 pts.) Pretend f is a magical function that has the property that at x = 3 the tangent line f is actually defined by the equation y = -2(x - 1) + 5. Obtain (a) f (3) = -2(3 - 1) + 5 = 1 (b) f (3) = -2

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

View Full Document
TEST2/MAC2311 Page 2 of 4 ______________________________________________________________________ 4. (25 pts.) Compute the first derivatives of the following functions. You may use any of the rules of differentiation that are at your disposal. Do not attempt to simplify the algebra in your answers. (a) f ( x ) 4 x 6 7 x 12 8tan( x ) f ( x ) 24 x 5 84 x 13 8sec 2 ( x ) (b) g ( x ) (4 x 2 2 x 1 )sec( x ) g ( x ) (8 x 2 x 2 )sec( x ) (4 x 2 2 x 1 )sec( x )tan( x ) (c) h ( t ) 5 t 10 1 sin( t ) 2 h ( t ) (50 t 9 )(sin( t ) 2) (5 t 10 1)cos( t ) (sin( t ) 2) 2 (d) y cot 5 (2 θ 1) dy d θ 5(cot(2 θ 1)) 4 ( csc 2 (2 θ 1))(2) 10cot 4 (2 θ 1) csc 2 (2 θ 1) (e) L ( z ) sin(4 z 8 ) 4 csc ( π 6 ) 4cos( z 2 ) dL dz ( z ) cos(4 z 8 )(32 z 7 ) 0 4( sin( z 2 ))( 1 2 ) 32 z 7 cos(4 z 8 ) 2sin( z 2 ) ______________________________________________________________________ Silly 10 point Bonus Problem: (Solution) Theorem 2.5.5 asserts that if g is a function with and f is a function that is continuous at L , then lim x c g ( x ) L lim x c f ( g ( x )) f ( L ). It follows from that the function f defined by lim x 0 sin( x ) x 1. is continuous at x = 0. Since sine is continuous and one-to-one on [- π /2, π /2], sin -1 is one-to-one and continuous on [-1,1], and thus, lim x 0 sin 1 ( x ) sin 1 (0) 0.
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