What is the equation of the line normal to # f(x)=e^(2x)/(e^x+4)# at # x=2#?

Question

What is the equation of the line normal to #  f(x)=e^(2x)/(e^x+4)# at #  x=2#?

Charles Anctil · Accepted Answer

In point slope form it is
#y-e^4/(e^2+4) = (e^2+4)^2/ (e^6+8e^4) (x-2)#.

For #f(x)=e^(2x)/(e^x+4)# at #x=2#, we get #y=e^4/(e^2+4)#. The slope of the tangent line is given by: #f'(x) = ((2e^(2x))(e^x+4)-e^(2x)(e^x))/(e^x+4)^2 = (e^(3x)+8e^2x)/(e^x+4)^2#. At the point where #x=2#, the slope of the tangent is #m_(tan) = (e^6+8e^4)/(e^2+4)^2#. So the slope of the normal line is #m_(norm) = (e^2+4)^2/ (e^6+8e^4)#. The equation of the line (in point slope form) through #(2,e^4/(e^2+4))# with slope #m = (e^2+4)^2/ (e^6+8e^4)# is #y-e^4/(e^2+4) = (e^2+4)^2/ (e^6+8e^4) (x-2)#. Rewrite using algebra as you see fit. (Or as your grader demands.)

William Abdalla · Answer

undefined To find the equation of the line normal to the function f(x) = e^(2x)/(e^x+4) at x = 2, we need to determine the slope of the tangent line at x = 2 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we can take the derivative of the function f(x) with respect to x and evaluate it at x = 2.

The derivative of f(x) = e^(2x)/(e^x+4) is given by:

f'(x) = (2e^(2x)(e^x+4) - e^(2x)(e^x))/(e^x+4)^2

Evaluating f'(x) at x = 2:

f'(2) = (2e^(4)(e^2+4) - e^(4)(e^2))/(e^2+4)^2

Now, we can find the slope of the tangent line by evaluating f'(2).

Next, we take the negative reciprocal of the slope of the tangent line to obtain the slope of the normal line.

Finally, using the point-slope form of a line, we can write the equation of the line normal to f(x) at x = 2.