Problems about maxima, minima, inflection points, asymptotes, ...

• READ THIS FIRST
  
• Problems about maxima, minima, inflection points, ...
  
  • Level 2 problems
    
  • Level 3 problems
    
• External link to solved problems on maxima and minima
  

READ THIS FIRST

If a problem is solved. It is not 'the' answer.
No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Problems about maxima, minima, inflection points, ...

Level 2 problems

• 
  

  What is the maximum and minimum value of cos(x)+sin(x)? Given : tan(pi.cos(x)) = cot(pi.sin(x)). What are the possible values of cos(x) + sin(x) ?

  
  

  1. Take y = cos(x) + sin(x), then y' = cos(x) - sin(x).

     y' = 0 <=> cos(x) = sin(x) <=> x = pi/4 + k.pi with k integer.

     Then y = sqrt(2) or y = -sqrt(2).

     The maximum and minimum values of cos(x)+sin(x) are sqrt(2) and -sqrt(2).

  2.  
           tan(pi.cos(x)) = cot(pi.sin(x))
     
     <=>   pi. cos(x) = pi/2 - pi.sin(x) + k.pi  with k integer.
     
     <=>   cos(x) + sin(x) = 1/2  + k  with k integer.
     

     Given the results of Part 1, the possible values of cos(x) + sin(x) are { 1/2 ; -1/2}.

• 
  

  Find the number of vertical asymptotes of f(x) = tan(x) + cot(x) in the interval [10,100].

  
  

  The asymptotes are the lines x = k.pi/2.

   
    k.pi/2 > 10    <=>    k > 20/pi   <=>    k > 6.37
    k.pi/2 < 100   <=>    k < 200/pi  <=>    k < 63.7
  

  The smallest k-value is 7 and the biggest is 63.
  So, there are 57 asymptotes in [10,100].

• 
  

  Prove that there is a real number r such that for all x _______ | a + x 1 x arctan( | ----- ) = - . arcsin(-) + r \| a - x 2 a The number a is a real strictly positive number. Then calculate the value of r.

  
  

   
  To prove this, we'll show that both sides have the same derivative.
  
                    _______
      d            | a + x         1           1/2         2a
      --   arctan( | ----- ) = -----------.-----------.-----------
      dx          \| a - x         a + x       _______         2
                               1 + -----      | a + x   (a - x)
                                   a - x      | -----
                                             \| a - x
  
                             1/2
                  = ... = -----------
                           _________
                          V a2 - x2
  and
  
      d    1          x          1/2 . 1/a
      -- ( - . arcsin(-) ) = ----------------
      dx   2          a       ____________
                             V 1- (x/a)2
  
  
             1/2
      =   ------------
           __________
          V a2 - x2
  
  

  Since the r-value is independent of x, we choose x = 0. Then it is easy to see that r = pi/4.

• 
  

  Calculate the condition for the real value of the parameter h such that the following function has a minimum and a maximum. h + 3x - x2 f(x) = --------------- x - 4

  
  

   
  f(x) is not defined for x = 4. Now we assume x is not 4.
  The derivative of the function is
                  - x2 + 8x - 12 - h
          f'(x) = -------------------
                     (x - 4)(x - 4)
  
          f'(x) = 0 <=>   - x2 + 8x - 12 - h = 0
  
          The discriminant D is 4(4-h)
  
          If D < 0 , or h > 4, f'(x) is negative for all x and there
          is no maximum and no minimum.
  
          If D > 0 , or h < 4, f'(x) has two roots and goes from
          negative to positive and to negative again.
          In that case, the function has a maximum and a minimum.
  

• 
  

  Recall previous problem and take h < 4. Investigate the number of intersection points of f(x) with the line y = k. Deduce from this the difference between the maximum and minimum value of f(x).

  
  
  The x values of the intersection points are the solutions of

   
          k(x - 4) = h + 3x - x2
  
  <=>     x2 + (k - 3)x - 4k - h
  
  From this, we see that there are at most two intersection points.
  The number of intersection points depends on the sign of
  
          D = (k - 3)2 + 16k + 4h
  
            = k2 + 10k + 9 + 4h
  
  D = 0 for k = -5 + 2.sqrt(4 - h) and for k = -5 - 2.sqrt(4 - h)
  
  For these k values there is just one intersection point.
  
  For k in ]-5 - 2.sqrt(4 - h); -5 + 2.sqrt(4 - h)[
  
  there are no intersection points.
  
  For k < -5 - 2.sqrt(4 - h) or k > -5 + 2.sqrt(4 - h)
  
  there are two intersection points.
  
  From this we deduce that the maximum value of f(x) =-5 + 2.sqrt(4 - h).
  
  The minimum value is -5 - 2.sqrt(4 - h).
  
  The difference between maximum and minimum is 4.sqrt(4-h).
  

• 
  

  Calculate the inflection points of y = (x + 1)/(x2 + 1)

  
  

   
                     -x2 - 2x + 1
          y' = ... = --------------
                      (x2 + 1 )2
  
                      2x3 + 6x2 - 6x - 2
          y" = ... = ---------------------
                          (x2 + 1 )3
  
  
                     2(x - 1)(x+2-sqrt(3))(x+2+sqrt(3))
          y" = ... = ----------------------------------
                          (x2 + 1 )3
  
  The inflection points are  (1;1)
  
                  (-2+sqrt(3);(1+sqrt(3))/4)
  
                  (-2-sqrt(3);(1-sqrt(3))/4)
  

• 
  

  Calculate the inflection points of y = sin(2x)+3sin(2x/3) in [0,3pi/2]. 3 hint : sin(3t) = 3.sin(t) - 4 sin (t)

  
  

   
          y' = 2cos(2x)+2cos(2x/3)
  
          y" = -4sin(2x) - (4/3)sin(2x/3)
  
          y" = 0 <=>   sin(2x) + (1/3)sin(2x/3) =0
  
          <=>     3.sin(2x/3) - 4sin3(2x/3) + (1/3)sin(2x/3) = 0
  
          <=>     sin(2x/3) . (10 - 12.sin2(2x/3)) = 0
  
          <=>     sin(2x/3) = 0 or sin2(2x/3) = 5/6
  
  The inflection points are (0;3pi/2) (1.725 ;2.43) and (2.986;2.43)
  

• 
  

  Calculate the inflection points of y = e2x - 5 ex +6

  
  

   
          y' =   2.e2x - 5 ex
  
          y" =   4.e2x - 5 ex
  
          y" = 0 <=> ex= 5/4  <=> x = 0.22
  
          Inflection point (0.223 ; 1.31)
  

• 
  

  Study the curve f(x) = (x2 - x)2 Give the number of intersection points of that curve with the line y = m.

  
  

   
          f'(x) = 2x(x - 1)(2x - 1)
  
          f"(x) = 2(6x.x -6x + 1)
  
          x       |       0       1/2     1
          ------------------------------------------
          f'(x)   |   -   0   +   0  -    0   +
          ------------------------------------------
          f(x)    |       0       1/16    0
  
  

  There is a minimum in (0,0) and (1,0) and a maximum in (3/2,1/16).
  f"(x) = 0 <=> x = 1/2 + 1/sqrt(12) or x = 1/2 - 1/sqrt(12)
  For these x-values, we have the inflection points.
  From the table, we see that the curve is never beneath the x-axis.
  So, the line y = m has no intersection points for m < 0.
  For m = 0, there are 2 intersection points.
  For 0 < m < 1/16 there are are 4 intersection points.
  For m = 1/16 there are 3 intersection points.
  For m > 1/16 there are 2 intersection points.
  The number of intersection points is the number of roots of the equation

   
          (x2 - x)2 - m = 0
  

• 
  

  f(x) = arcsin(tan(x/pi)) Find period of f(x) image of f(x) domain of f(x)

  
  

  1.  
       The period of arcsin(tan(x/pi))
     
       = the period of (tan(x/pi)
     
       = pi . ( the period of tan(x) )
     
       = pi . pi
     
       = pi2
     

  2. In each period-interval the image of (tan(x/pi) is R.
     So, the image of f(x) is equal to the image of arcsin(x).
     The image of f(x) is [ -pi/2 , pi/2 ].

  3.  
          x is in domain of f(x)
     <=>
         -1  = or <    tan(x/pi)  = or <  1
     <=>
         -pi/4 + k.pi  = or <  x/pi  = or <  pi/4 + k.pi
     <=>
        -pi2/4 + k pi2  = or <  x  = or <  pi2/4 + k.pi2
     

Level 3 problems

• 
  

  The curve F is a circle with radius r and center O. AB is a central line with points A and B on F. The line CD is parallel to AB with points C and D on F, such that ACDB is a isosceles trapezium ( |AC| = |DB| ). The angle(DOC) = 2t radians. Calculate the area of the trapezium as a function of t. Calculate the value of t such that this area is maximum.

  
  
  Take x-axis and y-axis orthogonal with origin O, such that A and B are on y. Then point C has coordinates (r.cos(t), r.sin(t)) .
  |AB| = 2r ; |CD| = 2r.sin(t) ; height = r.cos(t)
  The area of the trapezium is (2r + 2r.sin(t)).r.cos(t)/2
  = r².(cos(t) + sin(t).cos(t))
  

   
  The derivative = r2.( -sin(t) + cos2(t) - sin2(t) )
  
                  = r2.( -sin(t) + 1- 2.sin2(t) )
  
  Since t is in [0,pi/2], sin(t) is positive and the only value that
  makes the derivative change sign is sin(t) = 1/2 <=> t = pi/6.
  The trapezium is maximum for t=pi/6. Then the trapezium is half a regular
  hexagon.
  

• 
  

  A function y = f(x) is implicitly defined by y -x - sqrt(y.y + 2(x - 1)y + 4x) = 0 Calculate f(x) What is the domain of the function f Determine the maxima and minima

  
  

   
          y -x = sqrt(y.y + 2(x - 1)y + 4x)
  
  <=>     (y -x)2 = y2 + 2(x - 1)y + 4x   with  (y-x) >= 0
  
  <=>     ...
              x2 - 4x
  <=>     y = ---------   with y >= x
              4x + 2
  
          x is in domain of f
  
  <=>     x is not -1/2  and  y >= x
  
                                   x2 - 4x
  <=>     x is not -1/2  and       --------- - x >= 0
                                   4x + 2
  
  <=>     ...
  
                                  x(x + 2)
  <=>     x is not -1/2  and      ---------  =< 0
                                  (2x + 1)
  
  Investigation of sign gives
                  x |     -2      -1/2     0
                  -------------------------------
                    |  -   0   +    |   -  0   +
  
  So the domain f is ]-infty ; -2] U ]-1/2 ; 0 ]
                     (x + 2)(x - 1)
          y' = ... = ---------------
                     (2x + 1)(2x + 1)
  
  Investigation of sign gives
                  x |     -2      -1/2    1
                  -------------------------------
                 y' |  +   0   -    |  -  0  +
  
  So the function increases in ]-infty ; -2]
  and decreases in ]-1/2 ; 0 ].
  The image is relative maximum for x = -2. Then y = -2.
  The image is relative minimum for x = 0. Then y = 0.
  
  

• 
  

  Define all real values of m such that the asymptotes of the curve 2(m-1)x - m + 1 y = ---------------- (m+3)x + m intersect in a point above the line y = 2x-1

  
  

   
                                  -m
  The vertical asymptote is x = -------
                                  m+3
  
                                     2(m-1)
  The horizontal asymptote is  y = ----------
                                     m+3
  
  Now, we must have  y > 2x - 1
  
          2(m-1)    -2m
  <=>     ------ >  ---- -1
           m+3      m+3
  
  
  <=> ...
  
          5m + 1
  <=>   --------- > 0
          m + 3
  
  <=> (5m + 1)(m + 3) > 0
  
  The m values are defined by  ( m <-3  or m > 1/5 )
  

• 
  

  Investigate if there is a vertical asymptote at x = 0 for the function arcsin(2x) - 2 arcsin(x) y = ------------------------- x3

  
  

   
  We have to calculate the limit of f(x) for x -> 0
  
              arcsin(2x) - 2 arcsin(x)
          lim -------------------------
           0          x3
  
  This gives the case 0/0 . We use L'Hospitals rule.
  
                2 / sqrt(1 - 4 x2) - 1 / sqrt(1 - x2)
          = lim -----------------------------------------
             0              3 x2
  
                                              _______     _________
                                             |      2    |        2
                          2                 \| 1 - x  - \| 1 - 4 x
          = lim -------------------------. -----------------------------
             0     _________     _______
                  |        2    |      2              2
                3\| 1 - 4 x    \| 1 - x              x
  
  
  The first factor has limit 2/3 . On the second factor we use L'Hospitals rule.
  
  
                      -x /sqrt(1 - x2) + 4 x /sqrt(1 - 4 x2)
          = (2/3).lim -------------------------------------------
                   0                  2 x
  
  
  
                      -1 /sqrt(1 - x2) + 4 /sqrt(1 - 4 x2)
          = (2/3).lim -------------------------------------------
                   0                      2
  
  
          = 1
  
  
  
  So, there is no vertical asymptote at x = 0.
  

• 
  

  Calculate the horizontal asymptote to the function sin(1/x) y = ---------------- arctan(1/x)

  
  

   
  We have to calculate
  
                  sin(1/x)
           lim ----------------
          infty   arctan(1/x)
  
  This gives the case 0/0. We can use L'Hospitals rule, but first
  we can simplify the limit by the substitution
  
          1/x = t
  
  Then we have
  
                  sin(t)
        =  lim ------------
            0     arctan(t)
  
  Now, we use L'Hospitals rule.
  
                     cos(t)
          = lim --------------------  = 1
             0   1 / sqrt(1 + t2)
  
  So, y = 1 is a horizontal asymptote
  

External link to solved problems on maxima and minima