Calculus sin(x) function.

Question

On the graph of f(x)=6sin(6πx), 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

QUESTION: x), 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.

f(x) is at its lowest when sin(6πx) = -1, which is when the argument, 6πx = -π/2 + 2nπ.

i.e. 6x = -1/2 + 2n

x1 = -1/12 + n/3

f(x) is at its highest when sin(6πx) = 1, which is when the argument, 6πx = π/2 + 2nπ.

i.e. 6x = 1/2 + 2n

x2 = 1/12 + n/3

x1 and x2 are consecutive values for n = 0.

i.e. x1 = -1/12, x2= 1/12

f1 = f(x1) = -6,  f2 = f(1/12) = 6

Our lowest and highest points now are (x1,f1) = (-1/12, -6), (x2,f2) = (1/12, 6).

Slope between these two points is given by m = (f2 - f1) / (x2 - x1)

m = (6 - (-6)) / (1/12 - (-1/12)) = 12/(1/6) = 72

Slope = 72

Answer 2

Sine and cosine are periodic functions with period 2π, so that we evaluate the equation with period set between -π and π.  So, we have: f(x)=6sin(6πx), -π≦6πx≦π     The first and second derivative of function f(x) will be stated as follows respectively: Df(x)=36πcos(6πx) and D²f(x)=-216π²sin(6πx)

Since Df(x)=0 at extreme points, we have: 36πcos(6πx)=0 ⇒ 6πx=-π/2 and π/2    So, the locations of 2 extrema are x1=-1/12 and x2=1/12   Substitute x1 and x2 into D²f(x).  D²f(x1)=-216π²sin(-π/2)=216π²＞0,  D²f(x2)=-216π²sin(π/2)=-216π²＜0

Since Df(x1)=0 and D²f(x1)＞0, f(x1) is the minimum, and the value y1 is y1=f(-1/12)=6sin(-π/2)=-6   Thus, the coordinates of point P is P( -1/12, -6).   While, Df(x2)=0 and D²f(x2)＜0, so f(x2) is the maximum and the value y2 is y2=f(1/12)=6sin(π/2)=6  Thus,the coordinates of point Q is Q(1/12,6).

Therfore, slope=(y2-y1)/(x2-x1)={6-(-6)}/{(1/12)-(-1/12)}=12/(2/12)=12*6=72

Answer: The slope of the line which passes through P and Q is 72.