What is the equation of the tangent line of #f(x)=x^2/e^(x+2) # at #x=-2#?

Question

What is the equation of the tangent line of #f(x)=x^2/e^(x+2)  # at #x=-2#?

Christopher Agresta · Accepted Answer

#y=-8x-12# Writing #f(x)=x^2e^(-x-2)# so we get by the product and chain rule #f'(x)=2xe^(-x-2)+x^2e^(-x-2)(-1)# #f'(x)=e^(-x-2)(2x-x^2)# and we get #f'(-2)=e^(2-2)(-4-4)=-8# and we have #y=-8x+n# since #f(-2)=4# we obtain #n=-12#

Axel Alvarez · Answer

undefined To find the equation of the tangent line of f(x) = x^2/e^(x+2) at x = -2, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) with respect to x, which is f'(x) = (2x - x^2)e^(x+2)/e^(x+2).

Next, we substitute x = -2 into f'(x) to find the slope of the tangent line at x = -2.

f'(-2) = (2(-2) - (-2)^2)e^(-2+2)/e^(-2+2) = -4e^0/e^0 = -4/1 = -4.

So, the slope of the tangent line at x = -2 is -4.

Now, we need to find a point on the line. We substitute x = -2 into f(x) to find the corresponding y-coordinate.

f(-2) = (-2)^2/e^(-2+2) = 4/e^0 = 4/1 = 4.

Therefore, the point on the tangent line at x = -2 is (-2, 4).

Using the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept, we can substitute the values we found to get the equation of the tangent line.

y = -4x + b.

Substituting the point (-2, 4) into the equation, we can solve for b.

4 = -4(-2) + b.

4 = 8 + b.

b = -4.

Therefore, the equation of the tangent line of f(x) = x^2/e^(x+2) at x = -2 is y = -4x - 4.