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
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?
2 answers
What does it mean asymptotic behavior?
0 answers