Q. 74 - Calculus

Question

Prove that $\ln x$ is increasing and concave down on its entire domain $(0,\infty )$.

Step-by-Step Solution

Answer

$\ln x$ is increasing and concave down on its entire domain $\left(0,\infty \right)$.

Step 1. Given information

We have to prove that $\ln x$ is increasing and concave down on its entire domain $\left(0,\infty \right)$.

Step 2. Proof of the question.

Consider the graph,

From the graph it is clear that the signed area under the graph of $f=\frac{1}{t}$ and $x$-axis is positive only on $\left(0,\infty \right)$.

Since, $\ln x$ is the antiderivative of $\frac{1}{t}$ where $t$ is a dummy variable.

So,

$$\begin{gathered} \frac{d}{dx}\ln x \\ =\frac{1}{x} \end{gathered}$$

Hence, for $x>0$ derivative of $\ln x$ is positive so it is increasing.

Finding the second derivative,

$$\begin{gathered} \frac{d^2}{d{x}^2}\ln x \\ =\frac{d}{dx}\left(\frac{1}{x}\right) \\ =-\frac{1}{x^2} \end{gathered}$$

$-\frac{1}{x^2}$ is less than $0$ so the function is concave down.

Therefore, $\ln x$ is increasing and concave down on its entire domain $\left(0,\infty \right)$.