Og Function: -cos(3x)
Derivative Function: 3 sin(3x)
The period for the function and its derivative is 2π/3 radians. Over one period, when is the derivative function equal to its original function? That is, when is f(x) = f '(x) over 0≤x≤2π/3? Round your final answer to three decimal places.
Reiny or anyone Please help me with step by step solution algebraically
So in effect we are solving
3 sin(3x) = -cos(3x)
sin(3x)/cos(3x) = -1/3
tan(3x) = -1/3 , but we know we are in quadrant II for the given domain, since the tangent is negative in quadrant II
set your calculator to RAD, find arctan (+1/3)
I get 3x = .32175
so 3x = .32175 OR 3x = π - .32175 = appr 2.819842..
x = .10725 OR x = .93995
remember that the period of tan is π, so by adding π to any
previous answer will yield a new answer, as long as we stay below 2π/3 (2.094...)
To find when the derivative function is equal to the original function over the interval 0 ≤ x ≤ 2π/3, we can set the two functions equal and solve for x algebraically.
Given:
f(x) = -cos(3x)
f'(x) = 3sin(3x)
Interval: 0 ≤ x ≤ 2π/3
Setting the functions equal to each other:
-f(x) = f'(x)
We can substitute the given functions:
-cos(3x) = 3sin(3x)
Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation:
-cos(3x) = 3sin(3x)
-cos(3x) = √(1 - cos^2(3x))
Squaring both sides:
cos^2(3x) = 1 - 9sin^2(3x)
Using the identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation as:
cos^2(3x) = 1 - 9(1 - cos^2(3x))
Expanding and simplifying:
cos^2(3x) = 1 - 9 + 9cos^2(3x)
10cos^2(3x) = -8
cos^2(3x) = -4/5
Since the range of cos^2(3x) is between 0 and 1, there is no real solution to this equation.
Therefore, over the interval 0 ≤ x ≤ 2π/3, there is no solution where the derivative function is equal to the original function.