Math 1501 B1C,B2C

I'm Tom Morley

• 1. Find all x such that | x^2 - 10 x | <= 16. (<= is less than or equal.).
  

  If | x^2 - 10 x | = 16, then either (x^2 - 10 x ) = 16 or - (x^2 - 10 x ) = 16. The roots of these two quadatic equations are (in order):

    
    5-sqrt(41) < 2 < 8 < 5 + sqrt(41)
    

    At x = 5, | x^2 - 10 x = 0| = | -25| = 25 >  16, 
    at  x = -3, x^2 - 10 x =  (-3)^2 - 10 *(-3) = 36 >  16
    at x = 15,  x^2 - 10 x = 75 > 16
    at x = 0 , x^2 - 10 x = 0 < 16
    at x = 10 x^2 - 10 x = 0 < 16
    

  Therefore, the x such that | x^2 - 10 x | <= 16 are exactly :

    
      [5-sqrt(41),2] union [8, 5+ sqrt(41)]
      

• 2. From the definition, compute the derivative of 1/sqrt(2x + 3).

      
      
      
      ( Note --> = limit)
      
      
        f(x + h ) - f(x)
        ----------------
              h
              
      = 1/sqrt(2(x+h) + 3) - 1/sqrt(2x+3)
        ---------------------------------
                    h
                    
       = 1/sqrt(2(x+h) + 3) - 1/sqrt(2x+3)  1/sqrt(2(x+h) + 3) + 1/sqrt(2x+3)
        ---------------------------------   ---------------------------------
                    h                       1/sqrt(2(x+h) + 3) + 1/sqrt(2x+3)
                    
     =        1/(2x+2h+3) - 1/(2x+3)
       --------------------------------------
        h (1/sqrt(2(x+h) + 3) + 1/sqrt(2x+3))
        
     
     
     =   -2h / ((2x+2h+3) (2x+3))
       -------------------------------------
       h (1/sqrt(2(x+h) + 3) + 1/sqrt(2x+3))
       
       
    =   -2 / ((2x+2h+3) (2x+3))
       -------------------------------------  ---> 
       (1/sqrt(2(x+h) + 3) + 1/sqrt(2x+3))   
       
     =   -2 / ((2x+3) (2x+3))
       -------------------------------------  ---> 
       (1/sqrt(2x + 3) + 1/sqrt(2x+3))   
       
     
      =       -2 / ((2x+3) ^2)
       ----------------------- 
              (2/sqrt(2x + 3) 
              
      
      sqrt(2x + 3)          -1
      ------------ = ---------------
      ((2x+3) ^2)      (2x + 3) ^(3/2)
      
      

• 3. Find all points on the curve x^3 /3 - 2 x ^2 + 7/2 x + 6 such that the tangent line is perpendicular to the line 3 y - 4 = -6x

  The slope of the line is -2, so we are trying to solve
  
  f'(x) = 1/2.
  
  This is  
  
  x^2 - 4 x + 7 /2 = 2,  whose solutions are:
  
  x = 1 and x = 3. The points are therefore (x,f(x)) which are:
  
  
  (1, 47/6)
  (3, 15/2)