Solutions Exam 4 (Applications of Differentiation) 1. a. Applying the Quotient Rule we compute the derivative function of f as follows:

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1 MAT 4 Solutions Eam 4 (Applications of Differentiation) a Applying the Quotient Rule we compute the derivative function of f as follows: f () = 43 e 4 e (e ) = 43 4 e = 3 (4 ) e Hence f '( ) 0 for = 0 or = 4 Moreover, these are the only two critical numbers of f since e > 0 for all real values of Since f '( ) 0 for < 0, f '( ) 0 for 0 < < 4, and f '( ) 0 for > 4, the function is decreasing on the intervals (, 0), (4, ), and increasing on the interval (0,4) This implies that f has a local minimum at = 0 equal to f(0) = 0 and a local maimum at = 4 equal to f(4) = e [Summarize all these findings in a first derivative sign table] b Applying the Quotient Rule again to f we compute the second derivative function of f as follows: f () = ( 4 3 )e (4 3 4 )e (e ) = e = ( )( 6) e Hence f ''( ) 0 for = 0, =, or = 6 Moreover, these are the only critical numbers of f since e > 0 for all real values of Since f ''( ) 0 for < 0, 0 < <, and > 6, the graph of f is concave up on the intervals (, 0), (0,) and (6, ) Since f () < 0 for < < 6, the graph of f is concave down on the interval (,6) As a result, the graph of f has inflection points at (, 6 96 e) (,65) and (6, ) (,3) e 6 [Summarize all these findings in a second derivative sign table] c

2 Applying the Chain Rule we get the derivative f () = ( 3) 3 = ( 3) 3 Hence, f (c) = (c 3) 3 Since f(4) f() = = 3, this implies the following: 4 4 3f (c) = f(4) f() (c 3) 3 = 4 (c 3)3 = 8 c 3 = c = Therefore, there can be no value of c in the interval (,4) such that 3f (c) = f(4) f() This, however, does not contradict the Mean Value Theorem since the function f is not continuous at = 3 3 Since f () = = as the two critical numbers of f = +, solving the equation f () = 0 yields = ± Now, evaluating f at both endpoints and critical numbers results in the following values: f( ) = tan ( ) = ( π 4 ) = π 057 f() = tan () = ( π 4 ) = π 057 f(3) = 3 tan (3) 050 From these computations, we conclude that the absolute maimum of f in the interval [,3] is equal to π 0 57 and absolute minimum of f in the interval [,3] is equal π Here the derivative function is given by dy = d 4 Thus the second derivative function is computed using the Quotient Rule as follows: d y = (4 )+( ) = 8+ 4 d (4 ) (4 ) = 8 4+ = ( (4 ) (4 ) ) Since this epression is always negative for < <, we conclude that the curve is concave down and has no inflection points in the interval (,)

3 5 A valid graph here must have the two horizontal asymptotes y = 4 (left end behavior) and y = 6 (right end behavior) to satisfy the it conditions It should also pass through the points (0, ) and (3,) Moreover, the function should be increasing everywhere and change from concave up to concave down at the inflection point (3,) in order to satisfy the first and second derivative conditions, respectively Since f (3) is undefined, the graph must finally have a vertical tangent line at (3,) 6 a Since ln as, we recognize the indeterminate form We can then apply l Hospital s rule as follows: ln ln ln Once again we can apply l Hospital s rule since we have the same indeterminate form This yields ln 0 b Since (csc cot) = ( cos ) = 0 0 sin sin 0 ( cos sin ), we recognize the indeterminate form 0 We can then apply l Hospital s rule as follows: 0 0 ( cos sin ) = 0 (sin cos ) = 0 = 0 c First, note that ln ln 0 e e 0 0 We recognize the indeterminate form 0 with this it By l Hospital s rule, we 0 then have

4 ln We thus conclude that e 0 e d First, note that ( e ) = (e e ln ) = e (ln e ) We recognize the indeterminate form with this it By l Hospital s rule, we then have (ln e ) = We thus conclude that ( e (e ) = e 0 = ) = 0 B The derivative of this polynomial function is given by f () = This derivative can be turned into the quadratic function f () = 0X + 5X + with the substitution X = 50 Since f () = 0 yields only negative solutions for X (check this), we conclude that there are no real values of that solve the equation f () = 0 this implies that the function f has no local etrema

5 ln B Let f( ) be the function defined for > 0 Then we need to show that f() < since this inequality implies that 5ln < 5 for all positive values of Applying the Quotient Rule to compute the derivative of f we get: ln ln f '( ) ln 4 3 So, for 0, f '( ) 0 ln 0 ln e e This is the only critical number of f Since f '( ) 0 for 0 e and f '( ) 0 for e, we conclude that the function is increasing on 0, e, decreasing on e,, and thus has an absolute maimum at e equal to f( e) 084 e Now, since values of 084 0, this proves the inequality 5ln < for all positive 5