First find the slope of the tangent line = y'
y' = 5/2 * x^(3/2) + 1/2 * x^(-3/2)
The line in question has slope m=4
So, you want y'=4
5/2 * x^(3/2) + 1/2 * x^(-3/2) = 4
let u = x^(3/2) We can rewrite that as
5u/2 + 1/2u = 4
Multiply by 2u
5u^2 + 1 = 8u
5u^2 - 8u + 1 = 0
u = 1.46332 or 0.13667
so, x = u^(2/3) = 1.2889 or .26532
Evaluate y for those values of x to get the actual points.
for the perpendicular line, the slope m = -2/3. Do the same steps.
Given that y=x^(5/2)- x^(-1/2) .
a) Find the coordinates where the tangent line is parallel to the line y-4x-4=0.
b) Find the coordinates where the tangent line is perpendicular to the line 2y-3x-12=0.
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