Find the point on the graph of y=√x between (1,1) and (9,3) at which....?
Find the point on the graph of y=√x between (1,1) and (9,3) at which the tangent to the graph has the same slope as the line through (1,1) and (9,3). Can you please explain the steps on how to find the answer? Thanks in advance! :)
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slope of line through points, m1 = (3-1/(9-1) = 2/8 = 1/4
y = √x, y' = (1/2)x^(-1/2) = 1/4
x^-1/2 = 1/2
x^(-1/2)*2 = (1/2)^2
1/x = 1/4,
x = 4, y = 2,
(4,2) is the point