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Let f be defined on an interval, and let x1 and x2 denote points in that interval. (a)f is increasing on the interval if f(x1)

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Presentation on theme: "Let f be defined on an interval, and let x1 and x2 denote points in that interval. (a)f is increasing on the interval if f(x1)

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2 Let f be defined on an interval, and let x1 and x2 denote points in that interval. (a)f is increasing on the interval if f(x1)f(x2) whenever x10 for every value of x in (a,b), then f is increasing on [a,b]. (b) If f(x)<0 for every value of x in (a,b), then f is decreasing on [a,b]. (c) If f(x)=0 for every value of x in (a,b), then f is constant on [a,b]. Definition. If f is differentiable on an open interval I, then f is said to be concave up on I if f is increasing on I, and f is said to be concave down on I if f is decreasing on I. If f is continuous on an open interval containing a value x0,and if f changes the direction of its concavity at the point(x0,f(x0)), then we say that f has an inflection point at x0, and we call the point(x0,f(x0)) on the graph of f an inflection point of f

3 Use the graph of the equation y=f(x) in the accompanying figure to find the signs of dy/dx and dy/dx and d 2 y/dx 2 at the points A,B, and C. dy/dx 0 dy/dx >0, d 2 y/dx 2 <0 dy/dx<0, d 2 y/dx 2 <0

4 Use the graph of f shown in the figure to estimate all values of x at which f has(a) relative minima, (b) relative maxima, and (c) inflection points 2B2Because the curve is turning negative to positive. 0 Because the curve is turning positive to negative 3 Because the slope of f is changing negative to positive

5 Find any critical numbers of the function g(x) = x2(x2 - 6) g (x) = (x2) (x2 - 6) + (x2)(x2 - 6) g (x) = 2x(x2 - 6) + (x2)(2x) g (x) = 4x3 - 12x. g (x) = 4x(x2 - 3). Remember that a critical number is a number in the domain of g where the derivative is either 0 or undefined. Since g (x) is a polynomial, it is defined everywhere. The only numbers we need to find are the numbers where the derivative is equal to 0, so we solve the equation 4x(x2 - 3) = 0. The solutions are These are the only critical numbers.

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