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)