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Need help with this calculus question. Keep answers exact and show all calculations!

Need help with this calculus question. Keep answers exact and show all calculations-example-1
User Jorvis
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1 Answer

2 votes

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)

__

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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Need help with this calculus question. Keep answers exact and show all calculations-example-1
User T S Taylor
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6.9k points