The Mean Value Theorem

Return to List of Lectures

Previous Lecture

• Questions?
• Mean Value Theorem
  • If f is continuous over [a,b] and differentiable over (a,b), then there is some c between a and b such that ( f(b) - f(a) ) / (b - a) = f′c)
  • Examples
    • f(x) = x² - 3x
      • a = 1, b = 4
      • ( f(b) - f(a) ) / (b-a) = 2 = f′(c)
      • f′(x) = 2x - 3
      • Solve 2x - 3 = 2
        • 2x = 5
        • x = 2.5
        • i.e., c = 2.5, so f′(2.5) = 2
      • Look at this with R
        

        > f = function(x) { x^2 - 3 * x}
        > f(1)
        [1] -2
        > f(4)
        [1] 4
        > curve( f(x), 1, 4 )
        > a = 1
        > b = 4
        > ( f(b) - f(a) ) / (b - a)
        [1] 2
        > curve(  2*x - 4, add=TRUE, col="purple")
        > df = function(x) { 2*x - 3}
        > df(2.5)
        [1] 2
        > f(2.5)
        [1] -1.25
        > curve( 2*x - 6.25, add=TRUE, col="red")

        
        
    • f(x) = x³/3 - x
      • a = -2, b = 2
      • f(b) = 2/3
      • f(a) = -2/3
      • b - a = 4
      • Average slope = ( 2/3 - -2/3) / 4 = 1/3
      • To find c…
        • Find f′(x) = x² - 1
        • Set f′(x) = 1/3 and solve for x
          • x² - 1 = 1/3
          • x = √(4/3) = ± 2/√3 = c
        • c = ± 2/√3
        • Some of this with R…
          

          > f = function(x) {x^3/3 - x}
          > curve( f(x), -2, 2 )
          > m = 1/3
          > curve( m * x, add=TRUE, col="purple")

          
          
  • Significance
    • If f′(x) = 0, then f is constant (i.e., f(x) = k for some constant k)
    • If f and g have the same derivative, then f and g differ only by a constant (i.e., f(x) = g(x) + c)
      
  • Problem set
• Next
  • Monotonicity, extrema, and mean value theorem
    • Read section 4.3

Next Lecture