Answer:
y = (4 -x)e^-2
Explanation:
You want an equation of the tangent line to the graph of f(x) = xe^−x at its inflection point.
Inflection point
The inflection point on a curve is the point where the second derivative is zero, where the curve changes from being concave downward to concave upward, or vice versa.
We can use the product rule to differentiate f(x):
(uv)' = u'v +uv'
f'(x) = 1·e^-x +x·(-1)(e^-x) = (e^-x)(1 -x)
Then the second derivative is ...
f''(x) = (-e^-x)(1 -x) +(e^-x)(-1) = (e^-x)(x -2)
The second derivative is zero where one of its factors is zero. e^-x is never zero, so we have ...
(x -2) = 0 ⇒ x = 2
The point of inflection occurs at x = 2.
Point-slope equation
The point-slope equation of the line with slope m through point (h, k) is ...
y -k = m(x -h)
For this problem, we have ...
m = f'(2) = (e^-2)(1 -(2)) = -e^-2
(h, k) = (2, f(2)) = (2, 2e^-2)
So, the equation of the tangent line is ...
y -2e^-2 = -e^-2(x -2)
In slope-intercept form, this is ...
y = (-e^-2)x +4e^-2
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Additional comment
We can rearrange the equation to ...
y = (4 -x)e^-2
Usually a tangent line touches the graph, but does not cross it. The tangent at the point of inflection necessarily crosses the graph.