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April 1, 2015

April 1, 2015

Posted by **weloo_volley** on Tuesday, May 8, 2012 at 6:18pm.

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] .

- math -
**Reiny**, Tuesday, May 8, 2012 at 7:04pmYou 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

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