SOLUTION: Find an equation for the tangent line to the following curve,
h(x) = 19x − 5 ln x
at the point where x = e.
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Question 1107576: Find an equation for the tangent line to the following curve,
h(x) = 19x − 5 ln x
at the point where x = e.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find an equation for the tangent line to the following curve,
h(x) = 19x - 5ln(x)
at the point where x = e.
--------------
h(x) = 19x - 5ln(x)
h'(x) = 19 - 5/x
h'(e) = 19 - 5/e = slope of the tangent line.
===============
h(e) = 19e - 5
tangent point = (e,19e-5)
=============
y-y1 = m*(x-x1)
y-19e+5 = (19-5/e)*(x-e) = 19x - 19e - 5x/e + 5
y = 19x - 5x/e = x*(19 - 5/e)
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