Curve Sketching

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• Misc
  • Date for hour exam 2 could be as early as Nov. 11
    • Exam will cover material on derivatives since first exam (e.g., chain rule, implicit differentiation, related rates, extreme values, Mean Value Theorem, optimization, etc.)
    • Open reference materials
    • But harder, more word problems, than 1st test
• Questions?
• Monotonicity and Curve Sketching
  • Figure out where f(x) = (x²+1) / x is increasing, decreasing, (locally) maximum or minimum
    • How does f′(x) compare to 0 (less, greater, equal)?
    • f′(x) from quotient rule
      
• Concavity and Curve Sketching
  • Section 4.4
    • About concavity
    • Concave up means derivative is increasing on open interval
    • Concave down means derivative is decreasing
    • 2nd derivative test for concavity
      • 2nd derivative > 0 means concave up
      • 2nd derivative < 0 means concave down
    • Inflection point = point where function has tangent and concavity changes
    • 2nd derivative test for extrema
      • Derivative = 0, 2nd derivative < 0 means local max
      • Derivative = 0, 2nd derivative > 0 means local min
      • Derivative = 0, 2nd derivative = 0, test fails
    • Curve sketching
      • 2nd derivative tells you how 1st derivative is “curved” i.e., concavity
  • Examples
    • f(x) = x² - 3x + 1
      • f′(x) = 2x - 3
      • f′(x) = 0 implies x = 3/2
      • f(3/2) = -5/4 = -1.25
      • f′′(x) = 2
      • Tabulate derivative information and intervals to which it applies

        Interval	f′()	f′′()
        x < 3/2	-	+
        x = 3/2	0	+
        x > 3/2	+	+

      • 2nd derivative test says curve is concave up everywhere
      • So critical point is a minimum
      • Curve must look something like...
        
    • g(x) = x² + sin x
      • g′(x) = 2x + cos x
      • Find x such that 2x + cos x = 0
        • cos x = -2x
        • Calculator says x = -0.4502
        • Only one max/min! I intuitively expected a parabola with some oscillations superimposed by the sine function:
          
        • ...But this seems not to happen—x² presumably grows so fast it overwhelms the effects of sin x.
      • g′′(x) = 2 - sin x
        • 1 ≤ g′′(x) ≤ 3
        • So now we know that this function is always concave up, and that the critical point at -0.4502 is a minimum
  • Partial algorithm for curve sketching
    • Take derivatives
    • Find critical points f/ 1st derivative
    • Find intervals where 1st derivative is +, -
    • Find intervals where 2nd derivative is +, -, or 0
    • Find f(x) for key x values (critical points, x=0, etc.)
    • Find zeros of f(x)
• Next
  • See what more we can learn about x² + sin x
  • Optimization
    • Read section 4.5

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