f(x)=(2x-1)/e^x at x=1/2
Solve for the point (x_1, y_1)
let x_1=1/2
y_1=f(x_1)=(2x_1-1)/e^(x_1)
y_1=f(1/2)=(2(1/2)-1)/e^(1/2)
y_1=f(1/2)=(1-1)/e^(1/2)=0
y_1=0
Our point (x_1, y_1)=(1/2, 0)
Solve for the slope m
f(x)=(2x-1)/e^x
f' (x)=(e^x*d/dx(2x-1)-(2x-1)*d/dx(e^x))/(e^x)^2
f' (x)=((e^x*2)-(2x-1)*(e^x))/(e^x)^2
f' (x)=((2e^x-2xe^x+e^x))/(e^x)^2
Slope m=f' (1/2)=((2e^(1/2)-2(1/2)e^(1/2)+e^(1/2)))/(e^(1/2))^2
m=f' (1/2)=((2e^(1/2)-e^(1/2)+e^(1/2)))/(e)
m=f' (1/2)=(2e^(-1/2))
For the perpendicular line we need
m_p=-1/m=-1/((2e^(-1/2))
m_p=-1/2e^(1/2)
Solve for the equation of the line using color(blue)("Point-Slope Form")
y-y_1=m_p(x-x_1)
y-0=-1/2e^(1/2)(x-1/2)
color(red)(y=-1/2e^(1/2)(x-1/2))
Kindly see the graph of f(x)=(2x-1)/e^x (colored red) and the normal line y=-1/2e^(1/2)(x-1/2) (colored blue)
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God bless....I hope the explanation is useful.
