Posted by **maggie** on Friday, February 4, 2011 at 12:38am.

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.

- calculus - slope -
**MathMate**, Friday, February 4, 2011 at 6:23am
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.

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