This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 408N MIDTERM 1 Name: 9/22/2011 Bormashenko TA session: Show your work for all the problems. Good luck! (1) Let θ be an angle such that tan( θ ) = 1 2 and cos( θ ) < 0. (a) [5 pts] Calculate sec( θ ). (Please put this in the simplest form you can.) Solution: Recall that 1 + tan 2 ( θ ) = sec 2 ( θ ). Therefore, sec 2 ( θ ) = 1 + 1 2 2 = 1 + 1 4 = 5 4 Thus, sec( θ ) = ± r 5 4 = ± √ 5 2 Now we just need to figure out whether sec( θ ) is √ 5 / 2 or √ 5 / 2. Since cos( θ ) < 0, and sec( θ ) = 1 / cos( θ ), it must be true that sec( θ ) < 0. Therefore, sec( θ ) = √ 5 2 (b) [5 pts] Calculate sin( θ ). (This should also be put in simplest form.) Solution: This one can be done using part (a). Since tan( θ ) = sin( θ ) cos( θ ) = sin( θ ) · sec( θ ) by moving things to the other side, we get sin( θ ) = tan( θ ) sec( θ ) = 1 / 2 √ 5 / 2 = 1 √ 5 2 MATH 408N MIDTERM 1 (2) (a) [5 pts] Solve for x if 2 x +3 = 4 3 x 1 Solution: Writing everything as a power of 2, 2 x +3 = (2 2 ) 3 x 1 = 2 2(3 x 1) = 2 6 x 2 using exponent rules and expanding things out. To have powers of the same base be equal, the exponents have to be the same. Therefore, x + 3 = 6 x 2 ⇒ 5 = 5 x ⇒ x = 1 Thus, the answer is x = 1 . (b) [5 pts] Write the following quantity as a single logarithm: ln( x ) + ln( y 2 x ) 2ln( z ) Solution: Simplifying using logarithm rules, ln( x ) + ln( y 2 x ) 2ln( z ) = ln( x ) + ln( y 2 x ) ln( z 2 ) = ln x ( y 2 x ) z 2 MATH 408N MIDTERM 1 3 (3) The following is a graph of f ( x ): (a) [3 pts] Explain why f ( x ) is onetoone. Solution: f ( x ) is onetoone because it passes the horizontal line test. (b) [3 pts] Find the value of f 1 (3), explaining how you got it. Solution: f 1 (3) is the xvalue corresponds to y = 3. Therefore, f 1 (3) = 2 (c) [4 pts] Sketch the graph of f 1 ( x ) on these axes: Solution: Reflecting along the line y = x , we get: 4 MATH 408N MIDTERM 1 (4) Consider the f ( x ) in the following graph: (a) [5 pts] State whether lim x → 1 f ( x ) exists, and calculate it if it does....
View
Full
Document
This note was uploaded on 11/06/2011 for the course MATH 408N taught by Professor  during the Spring '11 term at University of Texas at Austin.
 Spring '11
 
 Math

Click to edit the document details