Find the equation of the tangent line to f(x)=1/sqrt x at x=4

Question

Please help me 

Answer 1

f(x) = 1/√x = x^-1/2

 

When x = 4, f(x) = f(4) = 1/√4 = 1/2

 

The point (4,1/2) is the point of tangency

 

f'(x) = (-1/2)x^-3/2 = (-1/2)[1/(√x)³] 

 

Slope of tangent line = (-1/2)[1/(√4)³] = (-1/2)(1/8) = -1/16

 

Equation of tangent line:  y-(1/2) = (-1/16)[x-4]

 

                                    y-(1/2) = (-1/16)x + 1/4

 

                                       y = (-1/16)x + (3/4)

 

To get another form of the equation without fractions, multiply through by 16:             16y = -x + 12

 

So, x + 16y = 12

 

Answer 2

The equation of a line is:

 

y = mx + b

 

where m is the slope and b is the y-intercept.  To answer this question we need to find the values of m and b.  The slope (m) is easy - it's just the value of the derivative of f(x) at x = 4:

 

     f(x) - x^-1/2

 

     f '(x) = -(1/2)x^-3/2

     f '(4) = (-1/2)4^-3/2 = -1/16 = m, the slope of the tangent line at x = 4

 

So the equation of the tangent line so far is:

 

     y = (-1/16)x + b

 

To find b, plug in the value of the known point (x=4, f(4)) that lies on the tangent line.

 

     f(x) = x^-1/2

     f(4) = 4^-1/2

     f(4) = 1/2

 

The known point is (4, 1/2)

 

    y = (-1/16)x + b

    1/2 = (-1/16)*4 + b

    1/2 = -1/4 + b

    3/4 = b

 

The final equation of the tangent line is:

 

     y= -(1/16)x + 3/4