Consider the function f(x)= -cos3x -4sin3x. (a)Find the equation of the line normal to the graph of f(x) when x= pie/6 . (b)Find the x coordinates of the points on the graph of f(x) where the tangent to the graph is horizontal. (c)Find the absolute extrema of the function f(x)=5+6x^3-3x^4 on the interval [-2,2] .
(a) To find the equation of the line normal to the graph of f(x) when x = π/6, we need to find the derivative of f(x) and then determine its value at x = π/6. The normal line will have a slope that is the negative reciprocal of the derivative at that point.
First, let's find the derivative of f(x). The derivative of each term can be found using basic differentiation rules:
f'(x) = -3sin(3x) - 4cos(3x) * 3
Simplifying gives:
f'(x) = -3sin(3x) - 12cos(3x)
Now, evaluate f'(x) at x = π/6:
f'(π/6) = -3sin(3(π/6)) - 12cos(3(π/6))
Since sin(π/6) = 1/2 and cos(π/6) = √3/2, we substitute these values:
f'(π/6) = -3 * (1/2) - 12 * (√3/2)
simplifying further gives:
f'(π/6) = - (3/2) - (6√3)
Now, we can find the slope of the line normal to the graph by taking the negative reciprocal of f'(π/6):
slope = -1 / ( -(3/2) - (6√3) )
Simplifying this expression yields:
slope = 2 / (3/2 + 6√3)
Now, we have the slope of the line normal to the graph of f(x) at x = π/6. Since we know the point of tangency is (π/6, f(π/6)), we can use the point-slope form of a line to find the equation of the normal line:
y - f(π/6) = slope * (x - π/6)
(b) To find the x-coordinates of the points on the graph of f(x) where the tangent to the graph is horizontal, we need to find the values of x when f'(x) = 0. In other words, we need to find the x-values where the derivative is equal to zero.
Taking the derivative f'(x) = -3sin(3x) - 12cos(3x) that we found in part (a), set it equal to zero:
-3sin(3x) - 12cos(3x) = 0
Now, we can solve this equation to find the x-values:
Divide both sides of the equation by -3 to simplify:
sin(3x) + 4cos(3x) = 0
Divide both sides of the equation by cos(3x) to isolate the tangent function:
tan(3x) = -4
Take the inverse tangent (arctan) of both sides to eliminate the tangent function:
3x = arctan(-4)
Finally, divide both sides of the equation by 3 to solve for x:
x = (1/3) arctan(-4)
(c) To find the absolute extrema of the function f(x) = 5 + 6x^3 - 3x^4 on the interval [-2, 2], we need to evaluate the function at the critical points inside the interval and at the endpoints.
First, find the critical points by finding where the derivative f'(x) = 0 or does not exist. Take the derivative of f(x):
f'(x) = 18x^2 - 12x^3
Set f'(x) equal to zero and solve for x:
18x^2 - 12x^3 = 0
Factor out an x:
x(18 - 12x) = 0
Setting each factor equal to zero, we find two critical points:
x = 0, and x = 3/2
Evaluate f(x) at the critical points and endpoints:
f(-2) = 5 + 6(-2)^3 - 3(-2)^4 = -59
f(0) = 5 + 6(0)^3 - 3(0)^4 = 5
f(2) = 5 + 6(2)^3 - 3(2)^4 = 29
Now, compare the function values obtained to find the absolute extrema:
The minimum value is -59 at x = -2, and the maximum value is 29 at x = 2.