To find the equation of a straight line #color(green)(y=color(red)ax+b)# we should have :
1) #color(red)(slope=a)#
2)Point #(color(brown)(x_1,y_1))# where the line passes through.
1)What is the #color(red)(slope)#
Since this line is normal and tangent to #f(x)# then the slope is by computing #color(red)(f'(-1))#
Since #f(x)# is a rational function of the form #(u(x))/(v(x))# then its derivative is:
#color(blue)(((u(x))/(v(x)))'=(u'(x)*v(x)-v'(x)*u(x))/(v^2(x))#
#f(x)=e^(4x)/(4x)#
#f'(x)=color(blue)(((e^(4x))'*(4x)-(4x)' (e^(4x)))/(4x)^2)#
#f'(x)=(4e^(4x)*4x-4e^(4x))/(16x^2)#
#f'(x)=(16xe^(4x)-4e^(4x))/(16x^2)#
Simplify by #4#
#f'(x)=(4xe^(4x)-e^(4x))/(4x^2)#
#color(red)(f'(-1))=(4(-1)e^(4(-1))-e^(4(-1)))/(4(-1)^2)#
#color(red)(f'(-1))=(-4e^(-4)-e^(-4))/4#
#color(red)(f'(-1))=(-5e^(-4))/4#
#color(red)(slope)# #color(red)( a=(-5e^(-4))/4)#
2)What is the coordinates of the point#(color(brown)(x_1,y_1))#?
Since the line is normal at #x=-1# so it passes through this point of abscissa #-1# let us find its ordinate #f(-1)#
#color(brown)(y_1)=f(-1)=(e^(-4))/-4#
#color(brown)(y_1)=-(e^(-4))/4#
Coordinates of the point are#(color(brown)(-1,-(e^(-4))/4))#
The equation of the normal line to #f(x)#is:
#y-color(brown)(y_1)=color(red)a*(x-color(brown)(x_1))#
#y-(-(e^(-4))/4)=(-5e^(-4))/4(x-(-1))#
#y+(e^(-4))/4=(-5e^(-4))/4(x+1)#
#y+(e^(-4))/4=((-5e^(-4))/4)x-(5e^(-4))/4#
#y=((-5e^(-4))/4)x-(5e^(-4))/4-(e^(-4))/4#
#y=((-5e^(-4))/4)x-(6e^(-4))/4#
Simplify #(6e^(-4))/4# by #2#
Therefore ,Equation of the line is:
#color(green)(y=((-5e^(-4))/4)x-(3e^(-4))/2)#
