find the tangent line

Question

find the tangent line to the function f(x)=x^1/2-1/x^1/2 at (1,0).

Answer 1

f(x) = x^1/2 - x^-1/2

Equation of line = mx+b, where m is slope and b is y-intercept.

To find the slope, m, take the derivative of f(x) then evaluate it at x = 1: m=df(1)/dx . Plug that into y=mx+b
To find b, plug the given point, (1,0), in for x and y in y = mx+b from step 1 and solve for b.

Answer 2

f'(x)= 1/2 x ^ -1/2 + x/2 ^ -3/2

 

f'(1) = 1/2(1)^(-1/2) + 1/2(1)^(-3/2)

 

f'(1) = 1/2 + 1/2

 

= 1

 

The slope is equal to 1.

 

0 = 1(1) + b

 

b = -1

 

y = x - 1

 

The equation of the tangent line would be given by y = x - 1

Answer 3

 

f(x) = x^1/2 - 1/x^1/2

 

Tangent at (1,0) is the derivative evaluated at (1,0)

 

—>  f'(x) = (1/2)x^-1/2 - (-1/2)x^-3/2 

        

       f'(x) = 1/(2x^1/2) + 1/(2x^3/2)

 

 

f'(1) = (1/2)(1) + (1/2)(1)  = 1  (= slope of f(x) at (1,0))

 

Tangent line is   y = mx + b

                        y = 1x + b

  We know the tangent line also passes through (1,0), use this to find b:

                        0 = 1 + b

                        b = -1

                     

  So the equation of the tangent line at (1,0) is:

                       y = x - 1 

 

 

 

Answer 4

f(x) = x^1/2 - x^-1/2

 

Equation of line = mx+b, where m is slope and b is y-intercept.

1. To find the slope, m, take the derivative of f(x) then evaluate it at x = 1: m=df(1)/dx .  Plug that into y=mx+b
2. To find b, plug the given point, (1,0), in for x and y in y = mx+b from step 1 and solve for b.