What is the equation of the normal line of #f(x)= (3x^2-2)/(6x)# at #x = 1#?

#f(x)=(3x^2-2)/(6x)# Let us first find the slope of the tangent. Slope of the tangent at a point is the first derivative of the curve at the point. so First derivative of f(x) at x=1 is the slope of the tangent at x=1 To find f'(x) we need to use quotient rule Quotient rule: #d/dx(u/v)=((du)/dxv-u(dv)/dx)/v^2# #u=3x^2-2=>(du)/dx=6x# #v=6x=>(dv)/dx=6# #f'(x)=((du)/dxv-u(dv)/dx)/v^2# #f'(x)=(6x(6x)-(3x^2-2)6)/(6x)^2# #f'(x)=(36x^2-18x^2+12)/(6x)^2##color(blue) "combine the like terms"# #f'(x)=(18x^2+12)/(36x^2)color(blue)"factor out 6 on the numerator "# #f'(x)=(6(3x^2+2))/(36x^2)color(blue)"cancel the 6 with the 36 in the denominator"# #f'(x)=(3x^2+2)/(6x^2)# #f'(1)=(3+2)/6=>f'(1)=5/6# #color(green)"slope of the tangent=5/6"# #color(green)"slope of the normal=negative reciprocal of slope of the tangent=-6/5"# #f(1)=(3-2)/6=>f'(1)=1/6# #color( red)" point-slope form of an equation of line "# #color( red) "y-y1=m(x-x1)...(where m:slope,(x1,y1) :points)"# We have slope =#-6/5 #and the points are #(1,1/6)# Use the point slope form #y-(1/6)=-6/5(x-1)=>y=(-6/5)x+6/5+1/6# #color(green)"combine the constant terms"# #color(green)"y=-6/5x+41/30"#