To find the slope of the tangent line, calculate f'(-2).
So f'(x) = e^(2x) + x*2e^(2x) = e^(2x)*(2x + 1), and thus f'(-2) = -3*e^(-4).
So the tangent line equation will be in the form y = (-3/e^4)x + b, where
x=-2 and y=xe^(2x) = -2e^(-4). Plug these values in to find b:
-2/e^4 = (-3/e^4)*(-2) + b ---> b = -2/e^4 - 6/e^4 = -8/e^4, so the equation is
y = (-3/e^4)x - 8/e^4.
So f'(x) = e^(2x) + x*2e^(2x) = e^(2x)*(2x + 1), and thus f'(-2) = -3*e^(-4).
So the tangent line equation will be in the form y = (-3/e^4)x + b, where
x=-2 and y=xe^(2x) = -2e^(-4). Plug these values in to find b:
-2/e^4 = (-3/e^4)*(-2) + b ---> b = -2/e^4 - 6/e^4 = -8/e^4, so the equation is
y = (-3/e^4)x - 8/e^4.
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Hey, Alba. e^(-4) = 1/e^4, and I prefer to write answers with positive exponents rather than negative exponents, so I divided by e^4 rather than multiplied by e^(-4). Just a personal preference, that's all. :)
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