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.

Answer 1

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=(y2-y1)/(x2-x1)
=(3-(-3))/(5π/4-3π4)
=3.82

Check my calculations.