4.3 1st and 2nd derivative tests

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1 CHAPTER 4. APPLICATIONS OF DERIVATIVES st and nd derivative tests Definition. If f 0 () > 0 we say that f() is increasing. If f 0 () < 0 we say that f() is decreasing. f 0 () > 0 f 0 () < 0 Theorem (First derivative test). To find the local ma/mins of a function f() do the following.. First find the critical points.. Figure out whether f 0 () is + or on each side of each critical point (four cases, lots of pictures): f 0 () =+, left of c right of c outcome + ) = c local ma + ) = c local min + + ) = c ) = c Eample. Draw a rough sketch of each possible kind of graph that we can have at a critical point. Solution. f 0 (c) =0 f 0 (c) DNE local ma f 0 () =+ local ma f 0 () =+ f 0 (c) =0 local min f 0 () = + f 0 (c) DNE local min f 0 () = +

2 CHAPTER 4. APPLICATIONS OF DERIVATIVES 09 f 0 (c) =0 f 0 (c) =0 f 0 (c) DNE () f 0 (c) DNE () f 0 () =++ f 0 () = f 0 () = f 0 () =++ f 0 (c) DNE (cusp) f 0 (c) DNE (cusp) f 0 () =++ f 0 () = This is where we ended on Friday, November 8 Finding when a function is positve or negative. To solve for when f() (or f 0 () or f 00 ()) is positive and when it s negative:. Solve for when f() (or f 0 () or f 00 ()) equals 0 or is undefined.. The numbers just found define intervals: each number is the endpoint of an interval. In other words, if we found a, b, c, d,...z, and if a<b<c< <z then the intervals are (,a) (a, b) (b, c)... (z,). 3. f() (or f 0 () or f 00 ()) is always positive or always negative on each interval from step : to find out which simply test it at any single value in the interval. Eample. Find the local ma/mins of f() = ln() p. Solution. Critical numbers f 0 () = p ln() / ( p ) = / ln() / = / ( ln()) = ln() / = ln() 3/

3 CHAPTER 4. APPLICATIONS OF DERIVATIVES 0 f 0 (0) DNE, but f(0) DNE too, so =0isnot a critical number. f 0 () =0 ln() 3/ =0 ln() =0 ln() = = e Plug numbers in on either side of each critical number The intervals are (0,e ) and (e, ). We want to pick numbers that are easy to plug into f 0 (): how about = e and = e? 0 (0,e ) f 0 (e) =+/? e (e, ) f 0 (e 3 )=+/? f 0 (e) = ln(e) = () = / = + + =+ f 0 (e 3 )= ln(e3 ) = (3) = / = + = Summing up f 0 () =+ at = e ) = e is local ma This is where we ended on Monday, November Eample 3. Find the local ma and mins of f() = /3 e 3. Solution.

4 CHAPTER 4. APPLICATIONS OF DERIVATIVES Critical numbers f 0 () = 3 /3 e 3 + /3 e 3 ( 3) = 3 e 3 3/3 /3 e 3 f 0 () DNE at = 0, and 0 is in the domain of f(), so = 0 is a critical point. f 0 () =0 3 /3 e 3 3 /3 e 3 =0 common factor e 3 3 /3 3 /3 =0 clear denominators, 3 /3 ) 3 /3 3 /3 3 /3 e 3 = 0 or 3 /3 3 /3 =0 e 3 is never 0 =3 /3 (0) 9 =0 = 9 Plug numbers in on either side of each critical number The critical numbers are 0 and /9, so the intervals are (, 0), (0, /9) and (/9, ). We want to pick numbers that are easy to plug into f 0 (): how about =, =/8 (chosen because I know how to take the cube root of 8) and =? (, 0) f 0 ( ) = +/? 0 (0, /9) f 0 (/8) = +/? /9 (/9, ) f 0 () = +/? f 0 ( ) = 3( ) /3 e3 3( ) /3 e 3 = e =+ = f 0 (/8) = 3(/8) /3 e 4 3(/8) /3 e 4 = e 4 3 / = e =+ +=+ f 0 () = 3() /3 e3 3() /3 e 3 = e =+ =

5 CHAPTER 4. APPLICATIONS OF DERIVATIVES Here s another way to figure out when f 0 () is positive and negative. Factor f 0 () f 0 () =e 3 3 /3 3 /3 = e 3 3 /3 9 /3 = e 3 ( 9) 3 /3 Note that e 3 is always positive (so is 3 ). Also, /3 is negative for <0 and positive for >0. Finally, 9 is positive for </9 and positive for >/9. Thus, we have e 3 ( 3 /3 9)=f 0 () <0 + + = 0 <</ = + /9 < + + = Summing up f 0 () = + at =0 ) = 0 is local min f 0 () =+ at =/9 ) = 0 is local ma Definition. If f 00 () > 0 we say that f() is concave up. If f 00 () < 0 we say that f() is concave down. f 0 > 0, f 00 < 0 incr, conc. down f 0 < 0, f 00 > 0 decr, conc. down f 0 > 0, f 00 > 0 incr, conc. up f 0 < 0, f 00 > 0 decr, conc up Interpretations of concavity: Concavity is not directly related to whether the graph is increasing or decreasing. Concave up means that the graph is curving more upwards or less downwards. Concave up means that the graph lies above each tangent line. Concave down means that the graph is curving more downwards or more upwards. Concave down means that the graph lies below each tangent line. Theorem (Second derivative test). To find the local ma/mins of a function f() try the following.. First find the critical points.

6 CHAPTER 4. APPLICATIONS OF DERIVATIVES 3. Figure out whether f 00 (c) is + or (three cases): f 00 (c) outcome + local min local ma 0 or DNE test says nothing Eample 4. Draw a rough sketch of each possible kind of graph that we can have at a critical point, and indicate what is happening with the second derivative. Solution. f 00 (c) = local ma f 00 (c) =+ local min f 00 (c) =0 local ma local min f 00 (c) =0 f 00 (c) =0 f 00 (c) =0 Eample 5. Repeat Eample, but use the second derivative test. Solution. We already know that f 0 () = ln(), and that the only critical 3/ point is = e. We take the second derivative, and plug in e : f 00 () = (0 )3/ ( ln()) 3 / ( 3/ ) = f 00 (e )= p 3 p ( ln()) 3 p p e 3 e ( ln(e )) (e ) 3 = e 3 e( ()) e 6 = e 3 e(0) e 6

7 CHAPTER 4. APPLICATIONS OF DERIVATIVES 4 So again = e is a local ma. = # 0 +# = # Definition. For any function f() we define. the intervals of increase/decrease: a list of the intervals where f() is increasing, and a list of the intervals where f() is decreasing.. the intervals of concavity: a list of the intervals where f() is concave up, and a list of the intervals where f() is concave down. 3. an inflection point: a number = c where the concavity changes from sign (i.e. the concavity is positive on one side and negative on the other). This is where we ended on Tuesday, November Eample 6. For the function f() = p, apple In each part below, you should find derivatives algebraically and solve for when they are zero or undefined algebraically. However, you may use your calculator to test when a derivative is positive or negative by plugging in a few values. (a) Apply the st derivative test. (b) Apply the nd derivative test. (c) Find the intervals of increase/decrease, intervals of concavity, and inflection points. (d) Use your calculator to graph the function, then label all of the information we have found. Solution. (a) Critical points f 0 () = p p Note that f 0 () DNE at =. However, this is an end point, and so we do not count it as a local ma/min. f 0 () =0 p p =0 clear denominators p p p = p (0) 4( ) = =0 ( 5 + 4) = 0 =0, 4/5

8 CHAPTER 4. APPLICATIONS OF DERIVATIVES Plug in numbers on either side of each critical point The intervals are (, 0) and (0, 4/5) and (4/5, ). We want to pick one -value in each interval, and calculate f 0 () there. Since we are using our calculators, it is easy to pick the values: let s use =, =/ and =0.9. Summing up f 0 ( ) 3. f 0 (/) 0.53 f 0 (0.9) 0.7 f 0 () = + at =0 ) = 0 is local min f 0 () =+ at =4/5 ) =4/5 is local ma (b) f 00 () = d p d = p + = p p p ( ) 4 p + p p 4( ) p p ( ) ( p ) simplify last fraction = p = p = p f 00 (0) = f 00 (4/5) p + p p 4( ) 4( )+ p 4( ) 3/ 3 +4 p 4( ) 3/ p p Summing up f 00 (0) > 0 ) = 0 is local min (again) f 00 (4/5) < 0 ) =4/5 is local ma (again) (c) Intervals of increase/decrease The intervals of increase/decrease were implied by our work on the first derivative test above: fis decreasing (, 0) [ (4/5, ) f is increasing (0, 4/5)

9 CHAPTER 4. APPLICATIONS OF DERIVATIVES 6 The intervals of concavity are found the same way as the intervals of increase and decrease. When is f 00 () = 0 or f 00 () DNE? Note that f 00 () DNE when =, but this doesn t matter since this is an endpoint. p 4( ) 3/ p f 00 () = p =0 4( ) 3/ 3 +4 p 4( ) 3/ = 4( ) 3/ (0) (clear denominators) 8( ) 4( ) ( 3 +4) =0 8( + ) = =0 4 +8=0 = 4 ± p 4 4()(8) () = 4 ± p = 4 ± 4p 6 30 = ± p 6 = p (because other # is > ) So, there are two intervals, p 6! and p! 6,. We pick an -value in each interval and evaluate f 00 () at that value. Actually, note that we ve already done this: the critical values =0, 4/5 and these values fall into the intervals. Recall f 00 (0) = f 00 (4/5) 4.47 thus! f is concave up, p 6 p! 6 f is concave down,.

10 CHAPTER 4. APPLICATIONS OF DERIVATIVES 7 Finally, we se that p 6 is the location of an inflection point because the concavity changes there from up to down. If we want the and y-values of the point, then we plug this -value back into f() f p! (d) The graph of f() isshownbelow: The red part of the graph shows when f() is increasing. The blue part shows when f() is decreasing. The arrows show when f() is concave down and and concave up. This is where we ended on Wednesday, November 3 Eample 7. [4.3#8] Sketch the graph of a function that has the following properties: f 0 () > 0 if <, f 0 () < 0 if > f 00 () < 0 if 0 <<3, f 00 () > 0 if >3 f 0 () = 0, f() =, f( ) = f(). lim! Solution. We start by drawing just a graph that is increasing and decreasing in the right regions, namely: f() is increasing for <<, and f() is decreasing for < and >.

11 CHAPTER 4. APPLICATIONS OF DERIVATIVES 8 Then we change this picture so that it has the right concavity, it flattens out at =, it has a horizontal asymptote, and it is odd (i.e. the left side is a certain kind of mirror image of the right side).