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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] .
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(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 at that point and then find the negative reciprocal of that slope to get the slope of the normal line.
1. Find the derivative f'(x) of the function f(x) = -cos(3x) - 4sin(3x).
- Differentiating -cos(3x) gives 3sin(3x) (using the chain rule).
- Differentiating -4sin(3x) gives -12cos(3x) (using the chain rule).
- So, f'(x) = 3sin(3x) - 12cos(3x).
2. Evaluate f'(π/6) to find the slope of the tangent line at x = π/6.
- Substitute x = π/6 into f'(x) to get f'(π/6).
- Calculate f'(π/6) = 3sin(3(π/6)) - 12cos(3(π/6)).
3. Find the negative reciprocal of the slope of the tangent line to get the slope of the normal line.
- Take the negative reciprocal of f'(π/6) to find the slope of the normal line.
Once you have the slope of the normal line, you can use the point-slope formula (y - y_1 = m(x - x_1)) to find the equation of the line that passes through the point (π/6, f(π/6)) and has the slope of the normal line.
(b) To find the x-coordinates of the points on the graph of f(x) where the tangent is horizontal, we need to find the values of x for which the derivative of f(x) is zero.
1. Find the derivative f'(x) of the function f(x) = -cos(3x) - 4sin(3x) (same as in part (a)).
2. Set f'(x) = 0 and solve the equation for x.
- Set 3sin(3x) - 12cos(3x) = 0 and solve for x.
The solutions to this equation will give you the x-coordinates of the points on the graph of f(x) where the tangent line 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 critical points and the endpoints of the interval and evaluate the function at those points.
1. Find the derivative f'(x) of the function f(x) = 5 + 6x^3 - 3x^4.
- Differentiate 6x^3 to get 18x^2.
- Differentiate -3x^4 to get -12x^3.
2. Set f'(x) = 0 and solve the equation for x to find the critical points.
- Set 18x^2 - 12x^3 = 0 and solve for x.
3. Evaluate the function f(x) at the critical points and the endpoints of the interval [-2, 2].
- Evaluate f(x) at the critical points and the endpoints of the interval to find the corresponding y-values.
The highest y-value will be the absolute maximum, and the lowest y-value will be the absolute minimum on the given interval.