(a) Find the slope m of the tangent to the curve y = 3/√x at the point where x = a > 0.(b) Find equations of the tangent lines at the points (1, 3) and (4,3/2) .

Answer 1

Hi Emanuel,

 y = 3/√x

First of all you need to find derivative of function

y' = (0 * √x - 3 *1/2√x) / (√x)^2

y' = ( -3/2√x) / x = -3/(2x√x)

slope at point x = a will be -3/ 2a√a

slope at point x = 1 will be -3 /2 = -1.5 then an equation at point (1,3) is:

y - 3 = -1.5 ( x - 1)

y = -1.5 x +4.5

slope at point x = 4 will be -3 / ( 2*4*√4) = -3/16 and an equation of tangent line at point (4, 3/2) is:

y - 3/2 = -3/16( x - 4 )

y = -3/16 x + 9/4

I hope it is useful