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
      • Secant line: y = mx + b = 2x + b
        • Goes through (1,-2)
        • (y-y0) = m(x-x0)
        • y + 2 = 2( x - 1) = 2x - 2
        • y = 2x - 4
      • Find c
        • Take derivative of f, set = (f(b) - f(a))/(b-a), solve for x
        • f′(x) = 2x - 3
        • 2x - 3 = 2, solve for x
        • x = 2.5 = c
      • Equation for line through (c,f(c)) with slope 2
        • y + 1.25 = 2(x - 2.5) = 2x - 5
        • y = 2x - 6.25
      • Visualize this in R
        

        > f = function(x) { x^2 - 3*x }
        > f(1)
        [1] -2
        > f(0)
        [1] 0
        > 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")
        > c = 2.5
        > f(c)
        [1] -1.25
        > curve( 2*x - 6.25, add=TRUE, col="red")

        
        
    • g(x) = (x³/3) - x
      • a = -2, b = 2
      • g(a) = (-2³ / 3 ) - -2
        • = -8/3 + 6/3 = -2/3
      • g(b) = ( 2³ / 3) - 2 = 8/3 - 6/3
        • = 2/3
      • ( g(b) - g(a) ) / (b-a)
        • = (2/3 - -2/3) / (2 - -2)
        • = (4/3) / 4
        • = 1/3
      • g′(x) = x² - 1
      • Solve for x in x² - 1 = 1/3
      • x² - 4/3 = 0 = (x - 2√3/3) ( x + 2√3/3)
      • Thus x = ± 2√3/3 = c
      • General form for tangent lines
        • y - y0 = m( x - x0 )
        • y0 = g(c)
        • x0 = c
        • y - g(c) = mx - mc
        • y = mx - mc + g(c)
      • Look at this function in R too
        

        > g = function(x) { x^3/3 - x}
        > curve( g(x), -2, 2 )
        > curve( x/3, add=TRUE, col="purple")
        > c = -2 / sqrt(3)
        > c
        [1] -1.154701
        > g(c)
        [1] 0.6415003
        > m = 1/3
        > t1 = function(x) { m*x - m*c + g(c) }
        > curve( t1(x), add=TRUE, col="red")
        > c = 2 / sqrt(3)
        > curve( t1(x), add=TRUE, col="red")

        
        
  • Significance
    • If f′(x) = 0, then f is constant (i.e., f(x) = k for some constant k)
    • Which implies that if f and g have the same derivative, then f and g differ only by a constant (i.e., f(x) = g(x) + c)
      • Which in turn means we “get it right” when we find anti-derivatives by simply asking “what function would differentiate into this?”
• Monotonicity, Extrema, and Mean Value Theorem
  • Section 4.3
    • Corollary 3: given f is continuous on [a,b] and differentiable on (a,b), then if f′(x) > 0 then f is increasing on [a,b]; if f′(x) < 0, f is decreasing on [a,b]
    • Monotonic = either always increasing or always decreasing
    • First derivative test for extrema
      • If f′ changes from - to + at c, then f has local minimum at c
      • If f′ changes from + to - at c, then f has local maximum at c
      • If f′ doesn’t change sign at c, then f has no extreme point at c
  • Examples: figure out where each of the following functions is increasing, decreasing, (locally) maximum or minimum
    • f(x) = 2x³ - x²
      • f′(x) = 6x² - 2x
        • = 2 (x) (3x-1)
      • Positive: f is increasing
        • Happens anywhere (x) and (3x-1) above are both positive or both negative
        • x > 1/3
        • x < 0
      • Negative: f decreasing
        • 0 < x < 1/3
      • 0: maximum/minimum
        • x = 1/3 (minimum, because f′(x) goes from decreasing to increasing
        • x = 0 (maximum, because f′(x) goes from increasing to decreasing
          
  • Problem set
• Next
  • Find increasing and decreasing intervals and extreme points of f(x) = (x² + 1) / x
  • Concavity
    • Read section 4.4

Next Lecture