weloo.
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] .
You are not "dumping" homework on us, are you?
I will do the first one, you do the others and let me know what you got
a)f(x) = -cos 3x - 4sin 3x
f'(x) = 3sin(3x) - 12cos(3x)
the slope of the tangent when x=π/6
= 3sin(π/2) - 12cos(π/2)
= 3 - 0 = 3
so the slope of the normal is -1/3
when x=π/6
f(π/6) = -cos π/2 - 4sin π/6 = 0 - 4 = -4
We need the equation of a line with slope -1/3 and a point (π/6, -4) on it
using y = mx + b
-4 = (-1/3)π/6 + b
b = π/18 - 4 or (π - 72)/18
y = (-1/3)x + π/18 - 4 or y = (-1/3)x + (π-72)/18
(a) To find the equation of the line normal to the graph of f(x) when x = π/6, we need to find the slope of the tangent line to the graph at that point and then use the negative reciprocal of that slope.
To find the slope of the tangent line, we need to take the derivative of the function f(x). The derivative of f(x) = -cos3x - 4sin3x can be found using the chain rule and trigonometric identities:
f'(x) = -3sin3x - 12cos3x.
Now, substitute x = π/6 into f'(x) to find the slope at x = π/6:
f'(π/6) = -3sin(3 * π/6) - 12cos(3 * π/6).
Simplifying this expression gives us:
f'(π/6) = -3sin(π/2) - 12cos(π/2) = -3(1) - 12(0) = -3.
So, the slope of the tangent line to the graph of f(x) at x = π/6 is -3.
The equation of a line in point-slope form is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. Since we want the line normal to the graph at x = π/6, we can use the point (π/6, f(π/6)).
Substituting the values into the equation, we have:
y - f(π/6) = -3(x - π/6).
To find f(π/6), substitute x = π/6 into the original function f(x):
f(π/6) = -cos(3 * π/6) - 4sin(3 * π/6) = -1 - 0 = -1.
Now, write the equation of the line normal to the graph at x = π/6:
y + 1 = -3(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 critical points of the function. Critical points occur where the derivative of the function equals zero or is undefined.
First, find the derivative of f(x):
f'(x) = -3sin3x - 12cos3x.
To find where the tangent is horizontal, we need to set the derivative equal to zero and solve for x:
-3sin3x - 12cos3x = 0.
Divide all terms by -3:
sin3x + 4cos3x = 0.
Now, we can use trigonometric identities to solve for x. One approach is to express sin3x and cos3x in terms of sinx and cosx:
3sinx - 4sin^3x + 4cos^3x - 3cosx = 0.
Rearrange and factor:
3(sin^2x - cos^2x) - 4sin^3x + 4cos^3x = 0.
Apply the Pythagorean identity sin^2x + cos^2x = 1:
3(1 - 2cos^2x) - 4sin^3x + 4cos^3x = 0.
Simplify:
3 - 6cos^2x - 4sin^3x + 4cos^3x = 0.
Combine like terms:
4cos^3x - 6cos^2x - 4sin^3x + 3 = 0.
At this point, we can solve this equation numerically using approximation methods or graphically to find the x-coordinates of the points where the tangent to the graph is horizontal.
(c) To find the absolute extrema of the function f(x) = 5 + 6x^3 - 3x^4 on the interval [-2, 2], we need to find the values of x within the interval that give the maximum and minimum values of f(x).
First, we find the critical points by taking the derivative of f(x):
f'(x) = 18x^2 - 12x^3.
To find the critical points, we set f'(x) equal to zero and solve for x:
18x^2 - 12x^3 = 0.
Factor out x^2:
x^2(18 - 12x) = 0.
Setting each factor equal to zero gives us:
x^2 = 0 or 18 - 12x = 0.
From the first equation, we have x = 0.
Solving the second equation, we get:
18 - 12x = 0
-12x = -18
x = 18/12
x = 3/2.
Now, we also need to consider the endpoints of the interval [-2, 2], which are -2 and 2.
Now, we evaluate f(x) at the critical points and the endpoints:
f(-2) = 5 + 6(-2)^3 - 3(-2)^4
= 5 + 6(-8) - 3(16)
= -37.
f(0) = 5 + 6(0)^3 - 3(0)^4
= 5.
f(2) = 5 + 6(2)^3 - 3(2)^4
= 5 + 6(8) - 3(16)
= 37.
To find the absolute extrema, we compare these values:
The minimum value of f(x) is -37 and it occurs at x = -2.
The maximum value of f(x) is 37 and it occurs at x = 2.
Therefore, the absolute extrema of the function f(x) on the interval [-2, 2] are a minimum at x = -2 with a value of -37 and a maximum at x = 2 with a value of 37.