Solve Tangent Lines Problems in Calculus

Tangent lines problems and their solutions, using first derivatives, are presented.

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 \( y = x^3 - 3 x \) below with the tangent lines.
  
  

Problem 2

• In order to find \( a \) and \( b \), we need to determine two algebraic equations in \( a \) and \( b \). The point of tangency is on the graph of \( y = a x^3 + b x \) and on the graph of \( y = - 3 x + 4\) at \( x = 1 \). Hence the y-coordinates of \( y = a x^3 + b x \) and \( y = - 3 x + 4\) at \( x = 1 \) are equal which after substitution of \( x = 1 \) gives the equation
  \( a \; (1)^3 + b \; (1) = - 3(1) + 4 \)
  
• Simplify the above equation in \( a \) and \( b \) to obtain
  \( a + b = 1 \)
  
• The slope of the tangent line is equal to \( -3 \) which is also equal to the first derivative \( y '\) of \( y = a x^3 + b x \) at \( x = 1\)
  \( y ' = 3 a x^2 + x = - 3 \) at \( x = 1 \).
  
• Substitute \( x = 1 \) in \( 3 a x^2 + x = - 3 \) to obtain a second equation in \( a \) and \( b \)
  \( 3 a + b = -3 \)
  
• Use any method to solve the system of equations \( a + b = 1 \) and \( 3 a + b = - 3 \) obtained above to get the solution:
  \( a = - 2 \) and \( b = 3 \).
  
• See graphs of \( y = a x^3 + b x \), 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 \( y = a \; e^x + b \; x \). Hence
  \( y ' = a \; e^x + b \)
  
• 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 is equal to zero)
  \( a e^x + b = 0 \)
  
• Rewrite the above equation as follows
  \( e^x = - b/a \)
  
• The above equation has solutions for \( -a / b \gt 0 \). Hence, the graph of \( y = a \; e^x + b \; x \) has NO horizontal tangent line if \( - a/b \le 0 \)

Exercises

Solutions to the Above Exercises