Can someone provide a step-by-step solution? Thanks!
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y = (x^2)e^-x
y' = (-e^-x)(x^2) + (2x)(e^-x) (product rule)
y' = e^-x(-(x^2) + 2x)
when x = 1,
y' = e^(-1)(-1 + 2)
y' = 1/e
Now, to find a tangent line, we use the y-intercept formula, plugging in our given x, y, and m;
y = mx+ b
1/e = (1/e)(1) + b
b = 0
So, the tangent line should be
y = (1/e)x
y' = (-e^-x)(x^2) + (2x)(e^-x) (product rule)
y' = e^-x(-(x^2) + 2x)
when x = 1,
y' = e^(-1)(-1 + 2)
y' = 1/e
Now, to find a tangent line, we use the y-intercept formula, plugging in our given x, y, and m;
y = mx+ b
1/e = (1/e)(1) + b
b = 0
So, the tangent line should be
y = (1/e)x