Find the slope of the tangent to the curve Find the slope of the tangent to the curve y=1/x^(1/2) at the point where x=a. Find the equations of the tangent lines at the points (1,1) and (4,1/2)
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First, differentiate y = 1/x^1/2.
==> y' = -1/[2x^(3/2)]
Substituing a for x, y' = -1/2a^(3/2) <---slope of the tangent at point x=a.
For the slope of the tangent line at (1,1), plug-in a = 1:
y'= -1/[2(1)^(3/2)] = -1/2
Using point-slope form,
y - 1 = -1/2(x - 1)
y - 1 = -1/2x + 1/2
y = -1/2x + 1/2 + 1
Equation of tangent line at point (1,1) is y = -1/2x + 3/2.
For the slope of the tangent line at (4,1/2), plug-in a = 4:
y'= -1/[2(4)^(3/2)] =-1/[2(8)] = -1/16
Using point-slope form,
y - 1/2 = -1/16(x - 4)
y - 1/2 = -1/16x + 1/4
y = -1/16x + 1/4 + 1/2
Equation of tangent line at point (4,1/2) is y = -1/16x + 3/4.
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