Problem 1

• Lines that are parallel to the x axis have slope = 0. The slope of a tangent line to the graph of y = x ³ - 3 x is given by the first derivative y '.
  y ' = 3 x ² - 3
  
• We now find all values of x for which y ' = 0.
  3 x ² - 3 = 0
  
• Solve the above equation for x to obtain the solutions.
  x = - 1 and x = 1
  
• The above values of x are the x coordinates of the points where the tangent lines are parallel to the x axis. Find the y coordinates of these points using y = x ³ - 3 x
  for x = - 1 , y = 2
  for x = 1 , y = - 2
  
• The points at which the tangent lines are parallel to the x axis are: (-1 , 2) and (1 , -2). See the graph of y = x³ - 3 x below with the tangent lines.
  
  

Problem 2

• We need to determine two algebraic equations in order to find a and b. Since the point of tangency is on the graph of y = a x³ + b x and y = - 3 x + 4, at x = 1 we have
  a(1)³ + b(1) = - 3(1) + 4
  
• Simplify to write an equation in a and b
  a + b = 1
  
• The slope of the tangent line is -3 which is also equal to the first derivative y ' of y = a x³ + b x at x = 1
  y ' = 3 a x² + x = - 3 at x = 1.
  
• The above gives a second equation in a and b
  3 a + b = -3
  
• Solve the system of equations a + b = 1 and 3 a + b = - 3 to find a and b
  a = - 2 and b = 3.
  
• See graphs of y = a x³ + b x, with a = - 2 and b = 3, and y = - 3 x + 4 below.
  
  

Problem 3

• The slope of a tangent line is given by the first derivative y ' of y = a e ^x + bx. Find y '
  y ' = a e ^x + b
  
• To find the x coordinate of a point at which the tangent line to the graph of y is horizontal, solve y ' = 0 for x (slope of a horizontal line = 0)
  a e ^x + b = 0
  
• Rewrite the above equation as follows
  e ^x = - b/a
  
• The above equation has solutions for -a/b >0. Hence, the graph of y = a e ^x + bx has NO horizontal tangent line if -a/b <= 0
  

Exercises

Solutions to the Above Exercises