Find the equation of the line tangent to

F(x) = xe^-x
at the point where x = 0. What does this tell you about the behavior of the graph when x = 0?

To find the equation of the tangent line to F(x) = xe^(-x) at the point where x = 0, we need to find the derivative of F(x) and evaluate it at x = 0.

Step 1: Find the derivative of F(x)
F(x) = xe^(-x)
To find the derivative, we can use the product rule.
We have two functions: f(x) = x and g(x) = e^(-x).
Using the product rule, we get:
F'(x) = f'(x) * g(x) + f(x) * g'(x)

f'(x) = 1 (derivative of x)
g(x) = e^(-x)
g'(x) = -e^(-x) (derivative of e^(-x) with respect to x)

Plugging in these values, we have:
F'(x) = 1 * e^(-x) + x * (-e^(-x))
= e^(-x) - xe^(-x)
= (1 - x)e^(-x)

Step 2: Evaluate the derivative at x = 0
To find the slope of the tangent line at x = 0, we evaluate F'(x) at x = 0.
F'(0) = (1 - 0)e^(-0)
= 1 * 1
= 1

The slope of the tangent line at x = 0 is 1.

Step 3: Find the equation of the tangent line
The equation of a straight line can be written in the form y = mx + b, where m is the slope and b is the y-intercept.

Since the slope of the tangent line at x = 0 is 1, we have m = 1. To find b, we need to find the value of y when x = 0.

F(0) = 0 * e^(-0)
= 0 * 1
= 0

The y-coordinate of the point where x = 0 on the graph of F(x) is y = 0.

Therefore, the equation of the line tangent to F(x) = xe^(-x) at the point where x = 0 is:
y = 1x + 0
y = x

This tells us that at x = 0, the graph of F(x) behaves like a straight line with a slope of 1.

To find the equation of the tangent line to a function at a given point, we need to determine the slope of the tangent line at that point and the coordinates of the point.

To find the slope of the tangent line, we can take the derivative of the function. Let's start by finding the derivative of F(x) = xe^(-x).

The derivative of F(x) with respect to x can be found using the product rule, which states that if u(x) = f(x)g(x), then the derivative of u(x) with respect to x is given by u'(x) = f'(x)g(x) + f(x)g'(x).

Applying the product rule, let f(x) = x and g(x) = e^(-x):

f'(x) = 1 (the derivative of x with respect to x is 1)
g'(x) = -e^(-x) (the derivative of e^(-x) with respect to x is -e^(-x))

Now, we can find the derivative of F(x):

F'(x) = f'(x)g(x) + f(x)g'(x)
= 1 * e^(-x) + x * (-e^(-x))
= e^(-x) - xe^(-x)

Next, we can find the slope of the tangent line at x = 0 by substituting x = 0 into the derivative:

F'(0) = e^(0) - 0 * e^(0)
= 1 - 0
= 1

Therefore, the slope of the tangent line at x = 0 is 1.

To find the coordinates of the point where x = 0, we substitute x = 0 into the function F(x):

F(0) = 0 * e^(-0)
= 0 * 1
= 0

Therefore, the point of tangency is (0, 0).

With the slope of the tangent line and the coordinates of the point, we can use the point-slope form of a linear equation to determine the equation of the tangent line. The point-slope form of a linear equation is given by y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

Substituting x1 = 0, y1 = 0, and m = 1 into the point-slope form, we get:

y - 0 = 1(x - 0)
y = 1x
y = x

Therefore, the equation of the tangent line to F(x) = xe^(-x) at the point where x = 0 is y = x.

This tells us that when x = 0, the graph of F(x) has a tangent line that is coincident with the line y = x. In other words, the tangent line and the graph of F(x) are the same at x = 0.

I did this for you on Tuesday.

http://www.jiskha.com/display.cgi?id=1365541972

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