How do you sketch the graph #y=ln(1/x)# using the first and second derivatives?

Question

Isaiah Allen · Accepted Answer

#f(x) = ln(1/x)# is monotone and strictly decreasing in its domain and therefore has no local extrema, and it is concave up everywhere.

Using the properties of logarithms we should see that: #ln(1/x) = -lnx# If we want to go through the whole sketcing process as an an exercise, we start by noting that #y=ln(1/x)# is defined and continuous for #x in (0,+oo)# and we analyze the limits at the boundaries of the domain: #lim_(x->0^+) ln(1/x) = lim_(y->+oo) lny = +oo# #lim_(x->+oo) ln(1/x) = lim_(y->0^+) lny = -oo# Then we calculate the first and second derivatives: #d/(dx) ln(1/x) = 1/(1/x) (-1/x^2) =-1/x# #d^2/(dx^2) ln(1/x) = 1/x^2# We can see that #f(x)# is monotone and strictly decreasing in its domain and therefore has no local extrema, and that it is concave up everywhere. graph{ln(1/x) [-10, 10, -5, 5]}

Charles Anctil · Answer

undefined To sketch the graph of $y = \ln(1/x)$ using the first and second derivatives, follow these steps:

1. Find the first derivative, $y' = \frac{d}{dx}[\ln(1/x)]$.
2. Simplify the first derivative.
3. Find the second derivative, $y'' = \frac{d^2}{dx^2}[\ln(1/x)]$.
4. Determine the critical points where the first derivative is zero or undefined.
5. Use the first and second derivatives to determine the behavior of the graph around critical points.
6. Sketch the graph based on the information gathered.

The first derivative is $y' = -\frac{1}{x}$ and the second derivative is $y'' = \frac{1}{x^2}$.

The critical point occurs at $x = 0$.

The first derivative is negative for $x > 0$ and positive for $x < 0$. The second derivative is positive for all $x$, indicating a concave up graph.

The graph approaches the x-axis as $x$ approaches infinity, and it approaches the line $y = 0$ as $x$ approaches zero from either side.