Question

What is the slope of the line normal to the tangent line of `f(x) = x^3-sqrtx` at `x= 3`?

Answer 1

The slope of tangent can be found by taking the first derivative. The slope of the normal is the negative reciprocal of the slope of the tangent.

Explanation:

`f(x) = x^3-sqrt(x)`
Slope of tangent `= f'(x) = 3x^2 - 1/(2sqrt(x))`
Slope of tangent at `x=3` is `f'(3) =3(3)^2-1/(2sqrt(3))`

Slope of tangent `= 27 - 1/(2sqrt(3)`

Simplifying we get `(54sqrt(3) - 1)/(2sqrt(3))`

The slope of normal would be the negative reciprocal of this value.

Slope of normal `= (-2sqrt(3))/(54sqrt(3)-1)`