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?
Have you graphed it? The graph ought to answer all the questions.
The tangent line will be
y=mx+b at x=2 then y=1/e^.5
if u= -x^-1
d(e^u)= e^u du/dx= e^-1/x * 1/x^2
so all that leads to
y=e^u
y'=1/x^2 * e^-1/x
for x=2, find y' the slope m, put it in the equation, and solve for b, and you have the tangent line.
Graph it all
What does it mean asymptotic behavior?
To graph the function f(x) = e^(-1/x) in the given window, you can follow these steps:
1. Set up a graphing tool or software, such as Desmos, and input the equation as y = e^(-1/x).
2. Set the x-axis range to -5 ≤ x ≤ 5 and the y-axis range to 0 ≤ y ≤ 10.
3. Plot the graph, which will show the curve of the function.
Now, let's find the tangent line to y = f(x) when x = 2:
1. Compute the derivative of f(x) to find the slope of the tangent line at any point x. The derivative of f(x) with respect to x can be found using the chain rule:
f'(x) = e^(-1/x) * (1/x^2)
2. Calculate the derivative f'(x) when x = 2:
f'(2) = e^(-1/2) * (1/2^2)
3. Simplify the expression:
f'(2) = e^(-1/2) * 1/4 = (e^(-1/2)) / 4
4. Now, we have the slope of the tangent line at x = 2. Plug this slope and the point (2, f(2)) into the point-slope formula to find the equation of the tangent line:
y - f(2) = (e^(-1/2)) / 4 * (x - 2)
5. Simplify the equation:
y = (e^(-1/2)) / 4 * x + (e^(-1/2)) / 2 - f(2)
Plot the tangent line on the graph, it should intersect the curve at x = 2.
To find the asymptotic behavior of the graph, we need to examine the behavior of the function as x approaches positive or negative infinity.
1. As x approaches positive infinity, the exponent -1/x approaches zero, making f(x) approach e^(0), which simplifies to 1. Therefore, the horizontal asymptote of the function is y = 1 as x approaches positive infinity.
2. As x approaches negative infinity, the exponent -1/x becomes increasingly large, resulting in the function's value approaching zero. Hence, the x-axis (y = 0) acts as a horizontal asymptote as x approaches negative infinity.
Regarding inflection points, we need to find where the concavity of the curve changes. To do this:
1. Compute the second derivative of f(x) by taking the derivative of f'(x) we found earlier:
f''(x) = (e^(-1/x) * (1/x^2))' = (-e^(-1/x) * (1/x^2)) / x^2
2. Since f''(x) includes e^(-1/x), which is never equal to zero for any x, there are no x-values for which the second derivative changes sign. Therefore, the graph of f(x) does not have any inflection points.
Include the asymptotes, the tangent line at x = 2, and the graph of f(x) on the same graph, and you will have a complete representation.