Calculus I Practice Test Problems for Chapter 2 Page 1 of 7

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1 Calculus I Practice Test Problems for Chapter Page of 7 This is a set of practice test problems for Chapter This is in no way an inclusive set of problems there can be other types of problems on the actual test The solutions are what I would accept on a test, but you may want to add more detail, and eplain your steps with words remember, I am interested in the process you use to solve problems! You should know the skills we reviewed in Chapter, for eample the quadratic formula, logarithm laws, eponential laws, algebra techniques (ie rationalizing, common denominator, epanding powers, etc), properties of basic functions (ie domain of logarithm, sketch of y, etc), functional notation (f(), then f(a + h) (a + h) ), etc There will be five problems on the test Most will involve more than one part You will have 00 minutes to complete the test You may not use Mathematica or calculators on this test Draw an eample of a function on the interval [a, b] for which the Intermediate Value Theorem does not apply Make sure your sketch is clearly labelled, and you eplain in words why the Intermediate Value Theorem does apply to your function Sketch a function g which has the following properties: g(), g has a removable discontinuity at, and g (0) Prove that if f and g are continuous at a, then f g is also continuous at a 4 Given f(), find f () using the definition of derivative 5 You are given a sketch of the function f below Use it to sketch the graph of the derivative f in the coordinate system provided Give a detailed eplanation about why you drew the curve f the way you did 6 Draw a well labelled sketch, and give a short eplanation of why the quantity representation of the slope of the tangent line to y f() at a You can assume h > 0 is yet another 7 A particle moves along a straight line with equation of motion s f(t) t t, where s is measured in meters and t in seconds Find the velocity when t

2 Calculus I Practice Test Problems for Chapter Page of 7 8 Find the horizontal and vertical asymptotes of the function: f() 9 Consider the function g() given below (c is a constant) Determine the value of c which makes g() continuous on (, ) g() { c, if < 4 c + 0, if 4 ( ) 0 Calculate the following using the it laws ( Calculate the following using the it laws + + ) Solutions Problem Answers to this question may vary The sketch of f() shows a function which has a jump discontinuity in the interval [a, b] Therefore, the Intermediate Value Theorem does not apply (since the function must be continuous) Problem Answers to this question may vary The sketch of g() has the desired properties Note that the slope of the tangent line to the curve at 0 is This is what the condition g (0) means

3 Calculus I Practice Test Problems for Chapter Page of 7 Problem If f g is continuous at a, we must have (f g)() (f g)(a) This is what we must show by our proof We are told that f and g are continuous at a, so we know that f() f(a) and g() g(a) ( ) (f g)() f() g() (algebra of functions) [ ] [ ] f() g() (it laws) [f(a)] [g(a)] (since f and g are continuous) f(a) g(a) (rewrite) (f g)(a) (algebra of functions) So since (f g)() (f g)(a), the function f g is continuous Problem 4 f f( + h) f() () ( + h) 0 indeterminant form; do some algebra h 0 ( + h) ( ( + h) + ) h ( + h) + ( ( + h) ) ( ) ( ( + h) + ) + h + h + h( ( + h) + ) h(h + ) h( ( + h) + ) h + ( + h) + now direct substitution will work 0 + ( + 0) + Problem 5 The major fact used in this solution is that the value of the derivative at a is equal to the slope of the tangent line to the curve at a The curve f() has a horizontal tangent at ±, therefore the derivative will be zero there That is, f () f ( ) 0 For ( 4, ) the function f() is decreasing (negative slope); therefore the derivative will be negative in this region That is, f () < 0

4 Calculus I Practice Test Problems for Chapter Page 4 of 7 For (, ) the function f() is increasing (positive slope); therefore the derivative will be positive in this region That is, f () > 0 For (, 4) the function f() is decreasing (negative slope); therefore the derivative will be negative in this region That is, f () < 0 Problem 6 We need a sketch that leads to Since we can assume h > 0, that means we will have a h < a on our sketch The slope of the line through P Q is h If we take the it as h approaches 0, the line through P Q approaches the tangent line at a, and therefore the tangent line to y f() at a has the slope m

5 Calculus I Practice Test Problems for Chapter Page 5 of 7 Problem 7 The velocity is equal to the derivative of the position f (a) f(a + h) f(a) f(t) t t f(a) a a f(a + h) (a + h) (a + h) (a + a h + ah + h ) a h a + 6a h + 6ah + h a h f (a) f(a + h) f(a) (a + 6a h + 6ah + h a h) (a a) [ a + 6a h + 6ah + h a h a + a] [6a h + 6ah + h h] h [ h(6a + 6ah + h )] (6a + 6ah + h ) Direct substitution now works! 6a + 6a(0) + (0) 6a Direct substitution won t work The velocity when t a s is v(a) f (a) 6a When t, the velocity is 6() m/s Problem 8 Horizontal Asymptotes: direct substitution does not work; indeterminant form / / / since / for > 0 + / 5/ So y / is a horizontal asymptote direct substitution does not work; indeterminant form / /

6 Calculus I Practice Test Problems for Chapter Page 6 of /( ) / + / 5/ since for < 0 So y / is a horizontal asymptote Vertical Asymptotes: The function might be infinite where the denominator is zero So 5/ is a possible vertical asymptote 5 + 5, since + > 0 and finite at 5/, and > 0 for 5/, since + > 0 and finite at 5/, and < 0 for 5/ There is a vertical asymptote at 5/ Problem 9 If a function g is continuous at a, then g() g() g(a) Our function g() is piecewise + defined For < 4, it is the polynomial c, so it is continuous (polynomials are continuous) For > 4 it is also a polynomial, so it will also be continuous in this region The only point we don t know if the function g() is continuous is at 4, not surprisingly the point where the definition changes We must choose c to make the function continuous at 4 We do this by by imposing that the following its be equal: g() g() Insert the proper definitions for g(): Evaluate by direct substitution: c(4) + 0 (4) c (c + 0) ( c ) A little algebraic rearranging gives us the following quadratic in c: c + 4c So if c satisfies this quadratic, then the left and right hand its will be equal The equality with g(a) that is required for continuity follows automatically in this case All that is left to do is solve the quadratic for c: c(4) + 0 (4) + c (c + ) 0 c So if c, the function g() will be continuous for R Problem 0 ( ) 0 0 direct substitution yields indeterminant quotient

7 Calculus I Practice Test Problems for Chapter Page 7 of 7 ( ) ( ) () 4 direct substitution will now work Problem ( + + ) direct substitution yields indeterminant difference ( ) ( ) rationalize the numerator direct substitution yields indetermninate quotient ( + ( ) ) Since > 0 (approaching infinity), we can say that + ( + + ) We can use the fact that 0 to get r ( )