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)

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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**.