Suppose f(x)= e^(-1/x). Graph this function in the window -5(</=)x(</=)5 and 0(</=)y(</=)10. Find the tangent line to y = f(x) when x=2 and include it in your graph. Also find any horizontal or vertical asymptotes and include them in your graph. Give explanations of the asymptotic behavior. Does the graph have any inflection points?

Question

Suppose f(x)= e^(-1/x). Graph this function in the window -5(</=)x(</=)5 and 0(</=)y(</=)10. Find the tangent line to y = f(x) when x=2 and include it in your graph. Also find any horizontal or vertical asymptotes and include them in your graph. Give explanations of the asymptotic behavior. Does the graph have any inflection points?

Answer 1

To graph the function f(x) = e^(-1/x) in the given window, we will first plot some points and then connect them to get a smooth curve.

Let's calculate a few points:
For x = -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, and 5, let's find the corresponding values of f(x).

For x = -5, f(x) = e^(-1/(-5)) = e^(1/5) ≈ 1.1487
For x = -4, f(x) = e^(-1/(-4)) = e^(1/4) ≈ 1.284
For x = -3, f(x) = e^(-1/(-3)) = e^(1/3) ≈ 1.3956
...
For x = 0, f(x) = e^(-1/0) = e^(-∞) = 0 (as x approaches 0, f(x) approaches 0).
...
For x = 5, f(x) = e^(-1/5) ≈ 0.8187

Now, let's plot these points on a graph with the given window. The x-values will be on the x-axis and the corresponding f(x) values will be on the y-axis.

Regarding the tangent line at x = 2, we need to calculate the derivative of f(x) to find the slope of the line. The derivative of f(x) with respect to x is given by:

f'(x) = (e^(-1/x))' = -(1/x^2)e^(-1/x)

Plugging in x = 2 into f'(x), we get:

f'(2) = -(1/2^2)e^(-1/2) ≈ -0.1839

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line, we substitute x1 = 2, y1 = f(2), and m = f'(2) to find the equation of the tangent line.

y - f(2) = -0.1839(x - 2)

Now let's plot this tangent line on the graph.

Regarding the asymptotic behavior, let's observe both horizontal and vertical asymptotes.

For the horizontal asymptote, as x approaches negative or positive infinity, f(x) approaches 0. Therefore, the horizontal asymptote is y = 0.

For the vertical asymptote, let's find the values of x for which the denominator of f(x) becomes zero, as it will result in an undefined value. The denominator is x, so we need to solve the equation x = 0. There is a vertical asymptote at x = 0.

For inflection points, we need to calculate the second derivative and find the x-values at which it equals zero. If there are any x-values that meet this criteria, the graph may have inflection points.