Asked by Rob
Find the tangent line approximation to 1/x when x=1
How do i find tangent line approximations??? my prof hasn't gone over this and i don't understand how to do it.
How do i find tangent line approximations??? my prof hasn't gone over this and i don't understand how to do it.
Answers
Answered by
Damon
y = 1/x
dy/dx = -1/x^2 = slope
at x = 1, slope = -1/1^2 = -1
find the line with that slope through that point (1,1)
1 = -1 *1 + b
b = 2
so line
y = -x + 2
dy/dx = -1/x^2 = slope
at x = 1, slope = -1/1^2 = -1
find the line with that slope through that point (1,1)
1 = -1 *1 + b
b = 2
so line
y = -x + 2
Answered by
bobpursley
The function is 1/x. There are a number of approximations, but I suspect you are using this one...
f(x+h)-f(x) / h
(1/(x+h)-1/x )/ h
so let h be some small increment, say .05
(1/1.05 - 1/1)/.05= you do it, that is the slope of the tangent line approximation.
f(x+h)-f(x) / h
(1/(x+h)-1/x )/ h
so let h be some small increment, say .05
(1/1.05 - 1/1)/.05= you do it, that is the slope of the tangent line approximation.
Answered by
Rob
thanks!!
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