Question

What is the equation of the normal line of `f(x)= 1-x-x^(1/3)` at `x=1`?

Answer 1

4y - 3x + 7 = 0

Explanation:

To find the equation of the normal we first must find the gradient of the tangent as the normal gradient is perpendicular to it.
To find m of tangent , obtain f'(x) and evaluate f'(1).

hence f'(x) = `-1 - 1/3x^(-2/3)`
and f'(1) = `- 1 - 1/3 = - 4/3 = " m of tangent "`
If 2 lines with gradients `m_1 " and " m_2 " are perpendicular"`

then `m_1 xx m_2 = -1`
here let `m_2 " be gradient of normal, and " m_1 " = m of tangent "`

then `m_2 xx -4/3 = -1 rArr m_2 = -1/(-4/3) = 3/4`
Now require a point on the normal line , knowing x = 1

hence f(1) = 1 - 1 -1 = -1 so point on line is (1 , -1 )

equation of normal: y - b = m(x - a ) , where (a,b)=(1,-1) and `m = 3/4`
so y + 1 = `3/4 (x - 1 )`

[multiply through by 4 to eliminate fraction]

4y + 4 = 3x - 3

`rArr 4y - 3x + 7 = 0`