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January 28, 2015

January 28, 2015

Posted by **LI** on Sunday, December 11, 2011 at 8:56pm.

The equation of the tangent line in slope-intercept form is y= ?

- calculus -
**Reiny**, Sunday, December 11, 2011 at 10:08pmI read that as

f(x) = e^(-2x) ln(8x)

f'(x) = -2(e^(-2x) ln(8x) + (1/x) e^(-2x) , using the product rule

f(2) = .0508

f'(2) = -.0924

so we have a slope of -.0924 and the point (2,.0508

y = -.0924x + b

sub in the point

.0508 = -.0924(2) + b

b = .2356

y = -.0924x + .2356

(Was expecting "nicer" numbers)

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