calculus
posted by maggie .
On the graph of f(x)=3sin(2x), points P and Q are at consecutive lowest and highest points with P occuring before Q. Find the slope of the line which passes through P and Q.

We will start with the general case of the function sin(x).
The minima (lowest points) of the graph of sin(x) occurs at
x=xmin=3π/2+2kπ where k∈ℤ (i.e. k=integer)
The maxima (highest points) of sin(x) occurs at
x=xmax=π/2+2kπ where k∈ℤ.
Two consecutive minimum/maximum could therefore occur at xmin=3π/2 and xmax=5π/2.
The given function is 3sin(2x), so
2x=3π/2, or x1= 3π/4 for minimum.
The ordinate at this point is
f(x1)=3sin(2*3π/4)=3
Therefore x1(3π/4,3).
and
2x=5π/2, or x2= 5π/4 for maximum.
The ordinate at this point is
f(x2)=3sin(2*5π/2)=3
Therefore x2(5π/4,3)
The slope is therefore
m=(y2y1)/(x2x1)
=(3(3))/(5π/43π4)
=3.82
Check my calculations.
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