Question

What is the equation of the tangent line of `f(x)=sqrtx` at `x=25`?

Answer 1

10y - x - 25 = 0

Explanation:

Rewrite the function f(x) = `sqrtx`

as f(x) = `x^(1/2)`

differentiating : `f'(x) = 1/2 x^(-1/2)`

Require the gradient (m) and the coordinates ( a , b ), a point on the line, to obtain the equation of the tangent.

m =f'(25 ) = `1/2 xx(25)^-(1/2) = 1/(2(25)^(1/2)) = 1/(2 xx 5) = 1/10`

and f(25) =`sqrt25 = 5`

Equation of tangent is y - b = m(x - a ) where m = `1/10`

and (a , b ) = (25 , 5 )

ie y - 5 = `1/10 (x - 25 )`

multiply both sides of the equation by 10 to eliminate fraction.

10y - 50 = x - 25 `rArr 10y - x - 25 = 0`