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=(y2-y1)/(x2-x1)
=(3-(-3))/(5π/4-3π4)
=3.82
Check my calculations.
To find the slope of the line passing through points P and Q on the graph of f(x) = 3sin(2x), the first step is to determine the x-coordinates of points P and Q.
Points P and Q occur at consecutive lowest and highest points of the function f(x) = 3sin(2x). The general formula for the x-coordinate of the highest and lowest points of the sine function is given by: x = (2n + 1)(π/2), where n is an integer.
Since P occurs before Q, we can take n = 0 for P and n = 1 for Q.
For P: xP = (2*0 + 1)(π/2) = π/2
For Q: xQ = (2*1 + 1)(π/2) = (3π/2)
Next, we need to find the corresponding y-coordinates for P and Q. We can substitute the x-coordinates into the function f(x) = 3sin(2x) to find the y-values.
For P: f(π/2) = 3sin(2(π/2)) = 3sin(π) = 0
For Q: f(3π/2) = 3sin(2(3π/2)) = 3sin(3π) = 0
Since both P and Q have y-coordinates of 0, the line passing through points P and Q is a horizontal line. The slope of a horizontal line is always 0.
Therefore, the slope of the line passing through P and Q is 0.