Question

DRG of 1/lnx

Answer 1

Domain: $(0, 1) \cup (1, \infty)$

Range: $(-\infty, 0) \cup (0, \infty)$

Graph: Vertical asymptote at $x=1$, horizontal asymptote at $y=0$. The graph is in two parts: for $0 < x < 1$ (below the x-axis) and for $x > 1$ (above the x-axis).

Answer 2

Explanation:

Domain:

1. The function involves $\ln x$, which is defined only for $x > 0$.

2. The function is a fraction with $\ln x$ in the denominator. The denominator cannot be zero, so $\ln x \neq 0$. This occurs when $x \neq e^0 = 1$.

3. Combining these conditions, the domain is $\{x \in \mathbb{R} \mid x > 0 \text{ and } x \neq 1\}$, which is $(0, 1) \cup (1, \infty)$ in interval notation.

Range:

Let $y = \frac{1}{\ln x}$. To find the range, we analyze the possible values of $y$.

As $x$ approaches the boundaries of the domain:

• As $x \to 0^+$ (from the right), $\ln x \to -\infty$, so $y = \frac{1}{\ln x} \to \frac{1}{-\infty} = 0^-$.

• As $x \to 1^-$ (from the left), $\ln x \to 0^-$ (small negative values), so $y = \frac{1}{\ln x} \to \frac{1}{0^-} = -\infty$.

• As $x \to 1^+$ (from the right), $\ln x \to 0^+$ (small positive values), so $y = \frac{1}{\ln x} \to \frac{1}{0^+} = +\infty$.

• As $x \to \infty$, $\ln x \to +\infty$, so $y = \frac{1}{\ln x} \to \frac{1}{+\infty} = 0^+$.

The function takes all values in $(-\infty, 0)$ and $(0, \infty)$. The value $y=0$ is never attained.

The range is $(-\infty, 0) \cup (0, \infty)$, or $\mathbb{R} \setminus \{0\}$.

Graph:

The graph of $f(x) = \frac{1}{\ln x}$ has the following features:

• Vertical Asymptote: At $x=1$, since $\ln x \to 0$ as $x \to 1$, causing $|f(x)| \to \infty$.

• Horizontal Asymptote: At $y=0$, since $f(x) \to 0$ as $x \to \infty$ and as $x \to 0^+$.

• Behavior: For $0 < x < 1$, $\ln x < 0$, so $f(x) < 0$. The graph is in the fourth quadrant, decreasing from near $(0,0)$ towards $-\infty$ as $x \to 1^-$. For $x > 1$, $\ln x > 0$, so $f(x) > 0$. The graph is in the first quadrant, decreasing from $+\infty$ as $x \to 1^+$ towards near $(0,0)$ as $x \to \infty$.