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

Feb 21, 2016

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) = $- 1 - \frac{1}{3} {x}^{- \frac{2}{3}}$
and f'(1) = $- 1 - \frac{1}{3} = - \frac{4}{3} = \text{ m of tangent }$
If 2 lines with gradients ${m}_{1} \text{ and " m_2 " are perpendicular}$

then ${m}_{1} \times {m}_{2} = - 1$
here let ${m}_{2} \text{ be gradient of normal, and " m_1 " = m of tangent }$

then ${m}_{2} \times - \frac{4}{3} = - 1 \Rightarrow {m}_{2} = - \frac{1}{- \frac{4}{3}} = \frac{3}{4}$
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 $m = \frac{3}{4}$
so y + 1 = $\frac{3}{4} \left(x - 1\right)$

[multiply through by 4 to eliminate fraction]

4y + 4 = 3x - 3

$\Rightarrow 4 y - 3 x + 7 = 0$