Anyone?
Please help me. I don't really understand this one much. Any help is extremely appreciated!
The Question:
The figure represents the graph of a derivative m'(x) for a function m(x).
Justify why for x=a and for x=g, the original graph shows local minima and there is a local maximum at x=e on the original graph of m(x)
The given figure is sinusoidal. It is a sine curve in that the start (x=a) and end (x=g) of one cycle are on the neutral/horizontal axis.
So if the given curve is that of the first derivative of the function m(x), then the m(x) is a cosine function.
m(x) = -cos(x)
m'(x) = -(-sin(x)) = sin(x)
Since cosX = sin(pi/2 -X), and the periods of sine and cosine curves are both 2pi, then the graph of basic cosX is one quadrant behind the graph of basic sinX.
In one cycle, or from 0 to 2pi, the basic sine curve has a local maximum and a local minimum. In the same 0 to 2pi, the basic cosine curve has only a local minimum and two local maxima, at the start and end. But if the cosine curve is negative, then it has only a local maximum and two local minima, at the start and end.
m(x) = -cos(x). A negative cosine curve. Only a local maximum in one period, and local minima at start and end.
In the given figure, the graph of the m(x)...
...starts at the lowest level at x=a, <----local minimum
...is at the neutral axis at x=b,
...is at the highest point at x=e, <----------the local maximum.
...is at the neutral axis again at x=f
...ends at the lowest level again at x=g. <----local minimum.
Hello,
there is an other attempt possible:
The function m' is at least of degree 3 (3 zeros). Thus the function m is at least of degree 4.
I've attached a sketch of m' and the corresponsing function m. I've labeled the numbers a, e, g as it is in your diagram. They are indicating the local extrema of the function m.
The extrema of m' indicate the points of inflection of the graph of function m.