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March 11th, 2014, 02:06 PM   #1
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calculating direction of tangent

a) Given the function g(x) = sin^2(x), what is the direction of its tangent at some x =xo?

I seem to be running into trouble with this one.

I have used a value of 5 for x.

using this formula y-f(xo)= f'(xo)(x-xo)

f'=2(sinx)(cosx)
sub for 5 and get -0.544

f(x) at 5 = 0.9195

Thus far I have

y-0.9195 = -.544(x-5)
y-0.9195= -0.544x + 2.72
y= -0.544x+3.64

When I plot this function I have a slope the looks around -.5 but why am I getting +3.64?
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March 11th, 2014, 07:41 PM   #2
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Re: calculating direction of tangent

That is where the tangent to at x = 5 intercepts the y-axis.

http://fooplot.com/plot/ooyzvg8bmj
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March 11th, 2014, 11:21 PM   #3
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Re: calculating direction of tangent

Thank you very much, makes perfect sense!
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