SOLUTION: find the equation for the line tangent in y=mx+b form. f(x)=sqrt (x)/5 at (4,2/5)

Algebra ->  Systems-of-equations -> SOLUTION: find the equation for the line tangent in y=mx+b form. f(x)=sqrt (x)/5 at (4,2/5)      Log On


   

Question 303783: find the equation for the line tangent in y=mx+b form. f(x)=sqrt (x)/5 at (4,2/5)
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+sqrt%28x%29%2F5
y' = %281%2F2%29x%5E%28-1%2F2%29%2F5
y' = 1%2F%2810sqrt%28x%29%29
At x = 4, y = 2/5, y' = 1/20 = m
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2/5 = (1/20)*4 + b
b = 1/5
--> y = (1/20)x + 1/5