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

##### 1 Answer

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 = 1hence 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 #