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**The Derivative in Graphing and Application**

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**Let f be defined on an interval, and let x1 and x2 denote points in that interval.**

f is increasing on the interval if f(x1)<f(x2) whenever x1<x2 f is decreasing on the interval if f(x1)>f(x2) whenever x1<x2 f is constant on the interval if f(x1)= f(x2) for all points x1 and x2. Let f be a function that is continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f’(x)>0 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

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**A: B: C: dy/dx<0 , d2y/dx2>0 dy/dx >0 , d2y/dx2<0**

Use the graph of the equation y=f(x) in the accompanying figure to find the signs of dy/dx and dy/dx and d2y/dx2at the points A,B , and C. A: dy/dx<0 , d2y/dx2>0 B: dy/dx >0 , d2y/dx2<0 C: dy/dx<0 , d2y/dx2<0

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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 A: Because the curve is turning negative to positive . B: 0 Because the curve is turning positive to negative C: 3 Because the slope of f is changing negative to positive

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**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). 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. Remember that a critical number is a number in the domain of g where the derivative is either 0 or undefined.

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