Let f be defined on an interval, and let x1 and x2 denote points in that interval. (a)f is...
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Transcript of Let f be defined on an interval, and let x1 and x2 denote points in that interval. (a)f is...
![Page 1: 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)](https://reader036.fdocuments.us/reader036/viewer/2022082805/551519d85503465e608b5044/html5/thumbnails/1.jpg)
![Page 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)](https://reader036.fdocuments.us/reader036/viewer/2022082805/551519d85503465e608b5044/html5/thumbnails/2.jpg)
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 x1<x2
(b) f is decreasing on the interval if f(x1)>f(x2) whenever x1<x2
(c) 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).
(a) 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
![Page 3: 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)](https://reader036.fdocuments.us/reader036/viewer/2022082805/551519d85503465e608b5044/html5/thumbnails/3.jpg)
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.
dy/dx<0 , d2y/dx2>0
dy/dx >0 , d2y/dx2<0
dy/dx<0 , d2y/dx2<0
![Page 4: 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)](https://reader036.fdocuments.us/reader036/viewer/2022082805/551519d85503465e608b5044/html5/thumbnails/4.jpg)
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
2 Because 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
![Page 5: 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)](https://reader036.fdocuments.us/reader036/viewer/2022082805/551519d85503465e608b5044/html5/thumbnails/5.jpg)
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 equation4x(x2 - 3) = 0.The solutions are These are the only critical numbers.
![Page 6: 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)](https://reader036.fdocuments.us/reader036/viewer/2022082805/551519d85503465e608b5044/html5/thumbnails/6.jpg)