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 #