Find an equation of the tangent line to the curve at the point (1, 1/e).
y = (x^3)(e^−x)
y = (x^3)(e^−x)
y' = -x^2 e^-x (x-3)
so, y'(1) = 2/e
Now you have a point and a slope, so the line is
y - 1/e = 2/e (x-1)
see
http://www.wolframalpha.com/input/?i=plot+y+%3D+%28x^3%29%28e^%E2%88%92x%29%2C+y%3D2%2Fe+%28x-1%29%2B1%2Fe+for+0.5+%3C%3Dx+%3C%3D+1.5
I took my derivative wrong when I did this on my own. Thank you!
To find the equation of the tangent line to the curve at the point (1, 1/e), we need to use the concept of derivatives. The derivative of a function gives us the rate of change of the function at a specific point.
Step 1: Find the derivative of the function y = x^3 * e^(-x).
To find the derivative, we can use the product rule. The product rule states that if we have a function of the form f(x) * g(x), then the derivative of the product is given by f'(x) * g(x) + f(x) * g'(x), where f'(x) and g'(x) are the derivatives of f(x) and g(x) respectively.
Let's solve for the derivative of y = x^3 * e^(-x):
Using the product rule, we have:
y' = (3x^2 * e^(-x)) + (x^3 * (-e^(-x)))
= 3x^2 * e^(-x) - x^3 * e^(-x)
Step 2: Substitute the values x = 1 and y = 1/e into the derivative equation.
To find the slope of the tangent line at the point (1, 1/e), we substitute x = 1 and solve for y'.
Let's substitute the values x = 1 and y = 1/e into y' = 3x^2 * e^(-x) - x^3 * e^(-x):
y' = 3(1)^2 * e^(-1) - (1)^3 * e^(-1)
= 3e^(-1) - e^(-1)
Step 3: Use the point-slope form to find the equation of the tangent line.
The point-slope form of the equation of a line is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
Using the point (1, 1/e) and the slope y' = 3e^(-1) - e^(-1), we can write the equation of the tangent line as:
y - (1/e) = (3e^(-1) - e^(-1)) * (x - 1)
Simplifying, we get:
y - (1/e) = 2e^(-1) * (x - 1)
And that is the equation of the tangent line to the curve y = x^3 * e^(-x) at the point (1, 1/e).