Chapter Three Chapter Three

Chapter Three Chapter Three

90 CHAPTER THREE ConcepTests for Section.. If f () = g (), then f() = g(). (a) True (b) False (b). If f () = g (), then f() = g() + C, where C is some constant. You might point out that the graphs of f and g differ b a vertical shift. A student s instinct is to answer True. Have students suggest their own countereamples.. If = π 5, then = 5π. (a) True (b) False (b). Since π 5 is a constant, then = 0. This question seems remarkabl obvious in class. However, on an eam students tend to miss this question.. If = ( + )( + )( + )( + ), then d5 d 5 = 0. (a) True (b) False (a). is a polnomial of degree four. Make sure our students realize that the don t need to take a derivative to answer this question.. The graph of a function f is given in Figure.. If f is a polnomial of degree, then the value of f (0) is (a) Positive (b) Negative (c) Zero f Figure. (b). Because the graph of this polnomial of degree is negative for large values of, the coefficient of will be negative. (Recall the third derivative of a polnomial of degree is a constant.) You could ask students wh f() could not become positive for >.

CHAPTER THREE 9 5. The graph of a function f is given in Figure.. If f is a polnomial of degree, then f (0) is (a) Positive (b) Negative (c) Zero f Figure. (a). Because the graph of this polnomial of degree is positive for large values of, the coefficient of will be positive. (Recall the third derivative of a polnomial of degree is a constant.) You could ask students if there could be other inflection points for f.. The graph of a function f is given in Figure.. If f is a polnomial of degree, then the values of f (0), f (0), and f (0) are (respectivel) (a) 0, 0, + (b) 0, 0, (c) 0, +, (d) 0,, (e) +,, + (f) 0, +, + f Figure. (f). There is a horizontal tangent at the origin, so f (0) = 0. The graph shows that f has horizontal intercepts at and 0, with a double root at 0. Thus f has the form f() = k( + ). Because f() > 0 for > 0, then k > 0. So f () = k( + ), f () = k( + ), and f () = k. You could ask students wh a double root at zero means that f has a factor of.

9 CHAPTER THREE 7. The graph of a function f is given in Figure.. If f is a polnomial of degree, then the values of f (0), f (0), and f (0) are (respectivel) (a) +, 0, + (b), 0, (c) +, 0, (d) +,, (e) +,, + (f) +, +, + 0.5 f 0.5 Figure. (c). The graph shows that f has horizontal intercepts at =, 0, and. Thus f has the form f() = k( + )( ). Because f() > 0 for 0 < <, we have k < 0. Then f () = k( ), f () = k(), and f () = k. You could ask wh there could not be another horizontal intercept outside this viewing window. 8. The graph of a function f is given in Figure.5. If f is a polnomial of degree, then the values of f (0), f (0), and f (0) are (respectivel) (a), +, + (b),, (c), +, (d), +, + (e) +,, + (f) +, +, + f 8 7 5 Figure.5 (c). At = 0, the graph is decreasing and concave up, so f (0) < 0 and f (0) 0. Because the graph becomes more negative as increases beond, the sign of the coefficient of (and thus the sign of f (0)) is negative. (Recall the third derivative of a polnomial of degree is a constant.) You could ask wh the graph of this function could not become positive for larger values of.

CHAPTER THREE 9 9. The graph of a function f is given in Figure.. If f is a polnomial of degree, then the values of f (0), f (0), and f (0) are (respectivel) (a),, + (b), 0, (c), +, (d), +, + (e) +,, + (f) +, +, + f Figure. (d). At = 0, the graph is decreasing and concave up, so f (0) < 0 and f (0) 0. Because the graph is positive as increases beond 0.5, the sign of the coefficient of (and thus the sign of f (0)) is positive. (Recall the third derivative of a polnomial of degree is a constant.) You could ask wh the graph of this function could not become negative for larger values of. ConcepTests for Section.. Figure.7 shows the graph of f() = a and its derivative f (). Which of the following are possible values for the parameter a? (a) a = (b) a = (c) a = (d) a = (e) None of the above f() = a f () Figure.7 (b) The onl possible value of a of the options listed is a =. The value a = is not possible, since the graph of f() = = is a horizontal line. Since the derivative of f() = a lies below the function, we know that a < e =.788... since the graph of the derivative of f() = e lies eactl on the graph of f() = e. The onl values of a that give a graph similar to the one shown (with the derivative below the function) are values of a between and e. Alternatel, we notice that ln a must be less than for the graph of the derivative f () = ln a a to lie below the graph of f() = a. Since ln > and ln >, the onl possible option is a =.

9 CHAPTER THREE ConcepTests for Section.. Given the graphs of the functions f() and g() in Figures.8 and.9, which of (a) (d) is a graph of f(g())? g() f() Figure.8 Figure.9 (a) (b) (c) (d) 0 8 (c). Because (f(g())) = f (g())g (), we see f(g()) has a horizontal tangent whenever g () = 0 or f (g()) = 0. Now, f (g()) = 0 for < g() < and this approimatel corresponds to.7 < <.5. You could have students give specific places in the other choices where there was a conflict with information given in the graphs of f and g.

CHAPTER THREE 95. Given the graphs of the functions f() and g() in Figures.0 and., which of (a) (d) is a graph of f(g())? g() f() 8 (a) Figure.0 (b) Figure. 5 8 (c) (d) 0 0 8 8 (a). Because (f(g())) = f (g())g (), we see f(g()) has a horizontal tangent whenever g () = 0 or f (g()) = 0. Now f () = 0 onl when =, so the composite function onl has horizontal tangents when g () = 0 or when g() =. Students ma want to check points, which is tedious. This is an opportunit to show the power of reasoning based on the chain rule.

9 CHAPTER THREE. Given the graphs of the function g() and f() in Figures. and., which of (a) (d) represents f(g())? 7 8 f() g() 5 9 9 8 7 Figure. Figure. (a) (b) 7 5 8 5 9 0 5 5 0 5 9 0 (c) 7 8 5 9 5 (d) 5 (c). Because (f(g())) = f (g())g (), we see f(g()) has a horizontal tangent whenever g () = 0 or g() = 0. This happens when = 0,, and. f(g()) is also negative for >. Alternativel, f(g()) = f(0) = 0 identifies answer (c). You could have students give specific places in the other choices where there was a conflict with information given in the graphs of f and g.

ConcepTests for Section. CHAPTER THREE 97. To differentiate the function ( + 5) +, begin b using (a) The product rule (b) The quotient rule (c) The chain rule (d) None of the above (b) The outside function here is a quotient of two inside functions, so we begin with the quotient rule. If our students can argue a wa to rewrite the epression so that one of the other rules is more appropriate, that would be great! You might ask them if there is an wa to rewrite the epression so that the chain rule is more appropriate, for eample, or the product rule. All are possible, depending on whether we rewrite the epression as ( + 5) for the chain rule, or for the product rule, or a variet of other possible was. + ( + ) ( + ) /. To differentiate the function + 5, begin b using (a) The product rule (b) The quotient rule (c) The chain rule (d) None of the above (c) The outside function here is the square root of a quotient, so the outside function is to raise something to the / power and we would begin with the chain rule. If our students can argue a wa to rewrite the epression so that one of the other rules is more appropriate, that would be great! You might ask them if there is an wa to rewrite the epression so that the quotient rule is more appropriate, for eample, or the product rule. All are possible, depending on whether we rewrite the epression as, for eample, + 5 for the quotient rule, or + (5 ) / for the product rule, or a variet of other possible was.. To differentiate the function begin b using (a) The product rule (b) The quotient rule (c) The chain rule (d) None of the above 5 +,

98 CHAPTER THREE (a) This is the product of two functions, so the product rule makes the most sense. If our students can argue a wa to rewrite the epression so that one of the other rules is more appropriate, that would be great! You might ask them if there is an wa to rewrite the epression so that the quotient rule is more appropriate, for eample, or the chain rule. All are possible, depending on whether we rewrite the epression as, for eample, (5 + ) for the chain rule, or (if the students reall get creative!) 5 + for the quotient rule, or a variet of other possible was.. To differentiate the function begin b using (a) The product rule (b) The quotient rule (c) The chain rule (d) None of the above ( + ) 5 +, (a) This is the product of two functions, so the product rule makes the most sense. If our students can argue a wa to rewrite the epression so that one of the other rules is more appropriate, that would be great! You might ask them if there is an wa to rewrite the epression so that the quotient rule is more appropriate, for eample, or the chain rule. 5. To differentiate the function it would be appropriate to use (check all that appl) (a) The product rule (b) The quotient rule (c) The chain rule f() = +, (b) and (c). If students recognize this function as ( + ), the chain rule is the best wa to find this derivative. ConcepTests for Section.5. For the function give the following four values: Amplitude of f(); Maimum value of f(); Amplitude of f (); Maimum value of f (). (a) ; 8; ; (b) 8; 8; ; (c) ; ; ; 0 (d) ; 8; ; (e) ; ; ; 8 f() = sin() + 5,

CHAPTER THREE 99 (d). The amplitude of f() = sin() + 5 is and the maimum value is 5 + = 8. The amplitude of f () = cos() is and the maimum value is.. List in order (from smallest to largest) the following functions as regards to the maimum possible value of their slope. (a) sin (b) sin() (c) sin() (d) sin(/) (d), (a), (b), (c). The derivative of sin(α) is α cos(α). The largest value of cos(α) is, so the ranking is based onl on the value of α. You could mimic this question replacing the sine function b cosine.. At which of the following values of does sin(π ) attain the largest slope? (a) = 0 (b) = / (c) = (d) = (d). The derivative of sin(π ) is 8π cos(π ). At these specific values of, we have the values 0, π, 8π, and π. You could mimic this question replacing the sine function b cosine.. Which of the graphs are those of sin() and sin()? (a) (I) = sin(), (II) = sin() (b) (I) = sin(), (III) = sin() (c) (II) = sin(), (III) = sin() (d) (III) = sin(), (IV) = sin() (I) (II) (III) (IV) (b). Because d sin(α) = α cos(α), the graph of sin() will have steeper slopes than sin(). The first positive d zeros are at = π/ for sin() and = π/ for sin(). You ma want to emphasize that there are man was of solving this problem, for eample, finding the -intercepts, remembering the properties of trigonometric functions, and using calculus to look at the slope of the function at various points. You could ask for equations for the graphs in choices (II) and (IV).

00 CHAPTER THREE 5. Which of the graphs is that of sin( )? (a) (b) 5 (c) (d) 5 π/ π π/ π (d). Since d d sin( ) = cos( ), the maimum value of the slope of the graph increases as increases. Also, the zeros of sin( ) are = nπ for n = 0,,,,..., which are closer together as increases. You could have students give specific points on the graphs in the other choices which have properties not consistent with those of = sin( ).