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(a) find an equation of the tangent line to the graph of at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your results.

Mathematics
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function: y= cos 3x point: (pi/4, -sqrt2/2)
where wer u stuck ?
i saw u doing a similar problem the other day ? :)

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Other answers:

take the derivative
I just get confused when trig is added in
do you know what the derivative is
dy/dx
-3sin3x
right
yes
so, slope = m = dy/dx = -3sin3x
since u want the slope at point: (pi/4, -sqrt2/2) evaluate slope at x = pi/4
dy/dx at x = pi/4 :- -3sin(3*pi/4) = ?
-0.123
that doesnt look right you should get :- dy/dx at x = pi/4 :- -3sin(3*pi/4) = \(\large \frac{-3}{\sqrt{2}}\)
now that u knw slope, m = \(\large \frac{-3}{\sqrt{2}}\) point: (pi/4, -sqrt2/2) wats the equation of tangent in point slope form ?
y-(-sqrt2/3)=(-3/sqrt2)(x-(pi/4)
?
thats right !
So that's for a? What would the graph look like for b? Do I just put that equation into Geogebra?
yup graph all 3 below :- 1) function: y= cos 3x 2) point: (pi/4, -sqrt2/2) 3) tangent : y-(-sqrt2/\(\color{red}{2}\))=(-3/sqrt2)(x-(pi/4)
u should get exactly like the one attached
1 Attachment

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