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The first derivative is set to zero to find the critical points of the function. A critical point can be a minimum, maximum, or a saddle point. There's a reason for this. Suppose a differentiable function f:R->R has a maximum at x=a. Then the function goes down to the right of a, which means f'(a)
GREEN'S THEOREM: if m=m(x,y) and n= n(x,y) are the continuous functions and also partial differential in a region 'r' of x,y plane bounded by a simple closed curve c. DIVERGENCE THEOREM: if f is a vector point function having continuous first order partial derivatives in the region v bounded by a closed curve s
Take a blank graph with 'x' and 'y' axes on it. Draw a 45-degree line on the graph. The line goes through the origin, and from the origin, it goes down-left and up-right. The slope of the line is 1, and its equation is y=x. The region "y is greater than or equal to x" is every point on that line, plus every point on the side above it (to the left of it).
y = - 6x2 - 12x - 1 A second degree equation graphs as a parabola, and has only one max or min. At that point, the first derivative y' = 0. dy/dx = - 12x - 12 = 0 - x - 1 = 0 ==> x = - 1 At that point, y = - 6( 1 ) - 12( - 1 ) - 1 = - 6 + 12 - 1 = 5. The max value of the function is 5, and occurs when x = -1.
the second derivative at an inflectiion point is zero