who needs a limit?
f = x + 4/x
f' = 1 - 4/x^2
So, f'(4) = 1 - 4/16 = 3/4
The tangent line at (4,5) is thus
y-5 = 3/4 (x-4)
using the limit
f(4+h)-f(4)
= (4+h)+4/(4+h) - (4 + 4/4)
= ((4+h)^2+4)/(4+h) - 5
= ((4+h)^2+4-5(4+h))/(4+h)
= (h^2+3h)/(4+h)
= h(h+3)/(h+4)
Now divide that by h and you get
(h+3)/(h+4)
as h->0, that -> 3/4
Find an equation of the tangent line to the graph of f at the given point.
f(x) = x + (4/x), (4,5)
I tried using the limit process a lot, but it didn't quite work. If possible, please use the limit process. Thanks a lot to whoever answers this question.
1 answer
1 answer