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MATH 1190 ñExam 1 (Version 2) Solutions September 7, 2007 S. F. Ellermeyer Name
Instructions. Your work on this exam will be graded according to two criteria: mathe- matical correctness and clarity of presentation. In other words, you must know what you are doing (mathematically) and you must also express yourself clearly. In particular, write answers to questions using correct notation and using complete sentences where appropriate. Also, you must supply su¢ cient detail in your solutions (relevant calculations, written explanations of why you are doing these calculations, etc.). It is not su¢ cient to just write down an ìanswerî with no explanation of how you arrived at that answer. As a rule of thumb, the harder that I have to work to interpret what you are trying to say, the less credit you will get. You may use your calculator on this exam but you may not use any books or notes.
- Write the pointñslope form equation of the line
y = 2x 5
that emphasizes the fact that the point (2; 1) is on this line. Solution: The equation of this line can be written in the pointñslope form
y ( 1) = 2 (x 2).
- (a) Graph the function f (x) = 4x^2 4 x + 3.
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
10 9 8 7 6 5 4 3 2 1 0
(b) Based on the graph that you have drawn, determine the intervals on which the function f is increasing and the intervals on which f is decreasing. You must write your answers in complete sentences using the correct terminology. For example, ìf is increasing on the interval (___; ___)î. Answers: f is decreasing on the interval ( 1; 0 :5) and f is increasing on the interval (0: 5 ; 1 ).
(c) Fill in the blanks:
- f has no relative maximums.
- f has a relative minimum value of 2 occurring at 0 : 5.
- Fill in the blanks:
If a function f is increasing on the interval (0; 2) and decreasing on the interval (2; 6) and if f (0) = 3 ; f (2) = 1 , and f (6) = 10 , then f has a relative maximum value of 1 occurring at 2.
- A graph of the function f (x) = 4x is shown below.
Since f (1) = 4, we can see that the point (1; 4) is on the graph of f.
(a) Fill in the following table which gives the slopes of certain secant lines. (Numbers can be rounded to six decimal places.) You will need your calculator to do these calculations, but you must show the steps of one sample calculation. (Also, be sure that your calculator is set to radians mode.) point (x; f (x)) slope of the secant line joining (1; 4) to (x; f (x)) (1: 1 ; 41 :^1 ) 5. (0: 9 ; 40 :^9 ) 5 : 177977 (1: 04 ; etc.) 5. (0: 96 ; _____________) 5. (1: 0052 ; ____________) 5. (0: 9948 ; _________________) 5. (1: 000072 ; ______________) 5.545454 (when rounded to 6 places) Show a sample calculation here: The slope of the secant line joining the point (1; 4) to the point (1: 1 ; 41 :^1 ) is
41 :^1 4 1 : 1 1
Show work here: This is a ìhardî limit problem, but we can convert it to an easy one by realizing that for all x 6 = 3, we have
x^2 3 x x^2 x 6
x (x 3) (x + 2) (x 3)
x x + 2
Thus lim x! 3
x^2 3 x x^2 x 6
= lim x! 3
x x + 2
- Find the derivative function, f 0 , of the function f (x) = 2x^2 5 x + 3. In doing this, you must use one of the two (equivalent) deÖnitions of the derivative, either
f 0 (x 0 ) = lim x!x 0
f (x) f (x 0 ) x x 0 or f 0 (x) = lim h! 0
f (x + h) f (x) h
and you must include all of the relevant algebraic details of your calculations. The answer that you should obtain is f 0 (x) = 4x 5. Solution:
f 0 (x) = lim h! 0
f (x + h) f (x) h
= lim h! 0
2 (x + h)^2 5 (x + h) + 3 (2x^2 5 x + 3) h
= lim h! 0
2 x^2 + 4xh + 2h^2 5 x 5 h + 3 2 x^2 + 5x 3 h
= lim h! 0
4 xh + 2h^2 5 h h = lim h! 0 (4x + 2h 5)
= 4x + 2 (0) 5 = 4x 5.
