1. f'=0 represents the vertex of the original equation because the slope(derivative) is zero, so it is not increasing or decreasing.
2. You can identify where a function increases when you calculate the derivative in front of the vertex and the point is less than the point of the vertex(or critical point). You can tell where it's decreasing if the derivative after the vertex is less than after the vertex(or critical point). If the derivative is negative, the equation is decreasing. If it is positive it is increasing.
3. The chain rule is the process of finding the derivative of a composite function.
If f(x)=g(h(x)) then f'(x)=g'(h(x))*h'(x)
In reality all your doing is...
1. Taking the derivative of the outside function
2. Rewriting the interior function
3. Multiplying by the derivative of the inside
Find the derivative and the equation of the tangent line where x=2.
f(x)= (2x^2 - x)^2 f'= 2(2x^2 - x) * 4x-1 = 2(4x-1)(2x^2 - x)
(8x^3 - 6x^2 + x)2 f'=16x^3 - 12x^2 + 2x
Point: (2, 36) Slope:84 Equation: y= 84x + 204
4. h(x)= f(g(x)) g(-4)=5 g'(-4)=2 f'(5)=20 Find h'(-4).
you could show how you got the slope and point for the tangent line on number three
ReplyDeleteYour method for determining whether a function f is increasing or decreasing is hard to follow. It shouldn't use vertices, because there are functions that do not have vertices, like the linear functions, for example.
ReplyDeleteKelvin makes a good point. Vertices are not necessarily present in the functions we will deal with. The most basic non-linear example would be the parent graph of a cubic. This has no vertex, but does have a possible max/min (critical point) at x = 0. Also, I cannot see your picture...
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