Answer:
y -1/e = -1/e(x -1)
Explanation:
You want the equation of the tangent line to the curve f(x) = x·e^(-2x +1) at x=1.
Tangent line
The equation of the tangent in point-slope form can be written as ...
y -k = m(x -h) . . . . . . . . . . point with slope m through point (h, k)
y -f(1) = f'(1)(x -1) . . . . . . . . point with slope f'(1) through point (1, f(1))
Slope
The slope of the line will be ...
f'(x) = e^(-2x +1) +x(-2)e^(-2x +1) = (1 -2x)e^(1 -2x)
f'(1) = (1 -2)e^(1 -2) = -1/e
Point
f(1) = 1·e^(1 -2) = 1/e
Equation
The equation of the tangent line in point-slope form is ...
y -1/e = -1/e(x -1)
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Additional comment
In slope-intercept form, it is ...
y = (-1/e)x +2/e
For tangent lines, we like the point-slope form because the equation is easily written from the derivative and the point of interest.
The point (1, 1/e) is also the point of inflection, so the tangent line crosses the curve at that point.
The relevant derivative forms are ...
d(uv) = du·v +u·dv
d(e^u) = (e^u)·du
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