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Applications of Differentiation. Chapter 5. Extrema on An Interval. Section 5.1. Definition of Extrema. Let f be defined on an interval I containing c . f(c) is the minimum of f on I if f(c) ≤ f(x) for all x in I .
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Applications of DifferentiationChapter 5Extrema on An IntervalSection 5.1Definition of ExtremaLet f be defined on an interval I containing c.
  • f(c) is the minimum of f on I if f(c) ≤ f(x) for all x in I.
  • f(c) is the maximum of fon I if f(c) ≥ f(x) for all x in I.
  • The minimum and maximum of a function on an interval are the extreme values, or extrema, of the function on the intervalDefinition of Extrema – Cont’dThe minimum and maximum of a function on an interval are the extreme values, or extrema, of the function on the interval.The minimum and maximum of a function on an interval are also called the absolute minimum and absolute maximum on the interval.The Extreme Value TheoremIf f is continuous on a closed interval [a,b], then f has both a minimum and a maximum on the interval.The minimum and maximum of a function on an interval are also called the absolute minimum and absolute maximum on the interval.Example – Finding Miminum and Maximum Values
  • f(x) = x2 – 4x + 5 on the closed interval [-1,3].
  • f(x) = x3 – 3x - 5 on the closed interval [-1,3].
  • Do they exist!Relative Extrema and Critical NumbersInformally, you can think of a relative maximum as occurring on a “hill” on the graph, and a relative minimum as occurring on a “valley” on the graph. Such a hill and valley can occur in two ways. If the hill (or valley) is smooth and rounded, the graph has a horizontal tangent line at the high point (or low point). If the hill (or valley) is sharp and peaked, the graph represents a function that is not differentiable at the high point (or low point)Definition of Relative ExtremaIf there is an open interval containing c on which f(c) is a maximum, then f(c) is called a relative maximum of f, or you can say that f has a relative maximum at (c,f(c)).If there is an open interval containing c on which f(c) is a minimum, then f(c) is called a relative minimum of f, or you can say that f has a relative minimum at (c,f(c)).ExampleFind the value of the Derivative at the Relative Extrema
  • f(x) = 9(x2 – 3)/x3
  • f(x) = x3 – 3x
  • Definition of Critical NumberLet f be defined at c: If f’(c) = 0 or if f is not differentiable at c, then c is a critical number of f.Relative Extrema Occur Only at Critical NumbersIf f has a relative minimum or relative maximum at x = c, then c is a critical number of f.Guidelines for Finding Extrema on a Closed Interval To find the extrema of a continuous function f on a closed interval [a,b], use the following steps.
  • Find the critical numbers of f in (a,b).
  • Evaluate f at each critical number in (a,b).
  • Evaluate f at each endpoint of [a,b].
  • The least of these values is the minimum. The greatest is the maximum.
  • Example – Find the Extrema on a Closed IntervalFind the Extrema on a Closed Interval.
  • f(x) = 3x4 – 4x3
  • f(x) = 2x = 3x2/3
  • f(x) = |1 – x2|
  • Rolle’s Theorem and the Mean Value TheoremSection 5.2Rolle’s TheoremLet f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a) = f(b)Then there is a least one number cin (a,b) such that f’(c) = 0ExampleFind the x-intercepts of f(x) = x2 – 3x + 2and show that f’(x) = 0 at some point between the two x-interceptsMean Value TheoremIf f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that f’(c) = [f(b) - f(a)]/(b-a)Example – Finding a Tangent LineGiven f(x) = 5 – (4/x), find all values of c in the open interval (1,4) such thatf’(c ) = [f(4)- f(1)]/(4-1) then the slope of the secant line is 1Now, find f’(x) = 4/(x2) = 1Solve for x, then x = 2 or -2, and c = 2 which is in the interval (1,4)Increasing and Decreasing Functions and the First Derivative TestSection 5.3Definitions of Increasing and Decreasing FunctionsA function f is increasing on an interval if for any two numbers x1 and x2 in the interval x1 < x2implies f(x1) < f(x2)A function f is decreasing on an interval if for any two numbers x1 and x2 in the interval x1 < x2implies f(x1) > f(x2)Theorem – Test for Increasing and Decreasing FunctionsLet f be a function that is continuous on the closed interval [a,b] and differentiable on the open interval (a,b)
  • If f’(x) < 0 for all x in (a,b), then f isdecreasing on [a,b]
  • If f’(x) > 0 for all x in (a,b), then f isincreasing on [a,b]
  • If f’(x) = 0 for all x in (a,b), then f isconstant on [a,b]
  • The First Derivative TestLet c be a critical number of a function f that is continuous on an open interval I containing c. If f is differentiable on the interval, except possibly at c, then f(c) can be classified as follows.
  • If f’(x) changes from negative to positive at c, then f has a relative minimum at (c,f(c))
  • If f’(x) changes from positive to negative at c, then f has a relative maximum at (c,f(c))
  • If f’(x) is positive or negative on both sides of c, then f(c) is neither a relative minimum or relative maximum
  • Concavity and the Second Derivative TestSection 5.4Definition of ConcavityLet f be differentiable on an open interval I. The graph of f is concave upward on I if f’ is increasing on the interval and concave downward on I if f’ is decreasing on the interval.Test for ConcavityLet f be a function whose second derivative exists on an open interval I.If f’’(x) > 0 for x in I, then the graph of f is concave upward in I.If f’’(x) < 0 for x in I, then the graph of f is concave downward in I.Definition of Point of InflectionLet f be a function that is continuous on an open interval and let c be a point in the interval. If the graph of f has a tangent line at the point (c,f(c)), then this point is a point of inflection of the graph of f if the concavity of f changes from upward to downward (or downward to upward) at the point.Finding points of Inflection
  • If (c,f(x)) is a point of inflection of the graph of f, then either f”(c) = 0 or f” does not exist at x = c.
  • Example:f(x) = x4 – 4x3f’(x)= 4x3 -12x2f”(x) =12x2 – 24xX = 0 and x = 2, so check the intervalsSecond Derivative TestLet f be a function such that f(c) = 0 and the second derivative of f exists on an open interval containing c.If f”(c)> 0, then f has a relative minimum at (c,f(c))If f”(c)< 0, then f has a relative maximum at (c,f(c))If f”(c) = 0, the test fails. That means f may have a relative maximum at or relative minimum at (c,f(c)) or neither. In that case use the first derivative test.Limits at InfinitySection 5.5Definition of Limits At InfinityUp to this point, most in explicit form. For example in the equation y = 3x2 - 5 the variable y is explicitly written as a function of xImplicit Form Explicit Form Derivativexy = 1 y = 1/x -1/x2Sometimes, this procedure doesn’t work, so you would use implicit differentiationImplicit DifferentiationTo understand how to find dy/dx implicitly, you must realize that the differentiation is taking place with respect to x. This means when you differentiate terms involving x alone, you can differentiate as usual, but when you differentiate terms involving y, you must apply the Chain Rule, because you are assuming that y is defined implicitly as a differentiable function of x.Differentiating with Respect to x
  • d/dx[x3] = 3x2
  • d/dx[y3] = 3y2dy/dx where y =u and n = 3
  • nun-1 u′
  • d/dx[x +3y] = 1 +3(dy/dx)
  • d/dx[xy2]= xd/dx +y2(d/dx)[x]
  • =2xy(dy/dx) + y2Guidelines for Implicit Differentiation
  • Differentiate both sides of the equation with respect to x.
  • Collect all terms involving dy/dx on the left side of the equation and move all other terms to the right side of the equation.
  • Factor dy/dx out of the left side of the equation.
  • Solve for dy/dx
  • ExampleFind dy/dx given that y3 +y2 -5y –x2 = - 4d/dx[y3 +y2 -5y –x2 = d/dx[- 4]3y2(dy/dx) +2y(dy/dx) -5(dy/dx) -2x =03y2(dy/dx) +2y(dy/dx) -5(dy/dx) =2xdy/dx(3y2 +2y – 5) = 2xdy/dx = 2x/(3y2 +2y – 5) Example – Finding the Slope of a Graph ImplicitlyFind the slope of the tangent line to the graph of 3(x2 + y2)2 = 100xy at the point (3,1)Find dy/dx and then use (3,1) to find the value of the slopeOther Applications of Implicit Differentiation
  • Finding a Differentiable Function
  • Finding the Second Derivative Implicity
  • Finding a Tangent Line to a Graph
  • Related RatesSection 4.6Guidelines for Solving Related-Rate Problems
  • Identify all given quantities and quantities to be determined. Make a sketch and label the quantities.
  • Write an equation involving the variables whose rates of change either are given or are to be determined.
  • Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time t.
  • Substitute into the resulting equation all known values for the variables and their rates of change. Then solve for the required rate of change.
  • ExampleAn airplane flying at an altitude of 6 miles is on a flight path that will take it directly over a radar tracking station. If distance s is decreasing at a rate of 400 miles per hour when s = 10 miles, what is the speed of the plane?Draw and label a right triangle. s is the hypotenuse of the right triangle and x is the horizontal distance from the station.Example – Cont’ds = 10x = (102 – 62)½ = 8ds/dt = -400 when s =10Find dx/dt when x=10 and x = 8The function is x2 + 62 =s2
  • Differentiate with respect to x
  • Solve for dx/dt
  • Substitute for s,x and ds/dt
  • ENDThe end
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