91Ó°ÊÓ

The given function is \(f(x)=\ln|x|\). The natural logarithm function is usually defined for positive values of x, but since we are using the absolute value, this function is defined for all non-zero real numbers.

We need to define the behavior of the graph of the function \(f(x)\) in the following cases: 1. When \(x=0\), the natural logarithm function is not defined; hence, we will have a vertical asymptote at \(x=0\). 2. When \(x>0\), the function is equivalent to \(f(x)=\ln x\). Here, as x approaches infinity, the natural logarithm function slowly increases, so the limits are: - \(\lim_{x \to 0^+} f(x) = -\infty\) - \(\lim_{x \to +\infty} f(x) = +\infty\) 3. When \(x<0\), the function becomes \(f(x)=\ln (-x)\). Here, as x approaches negative infinity, the natural logarithm function slowly increases, so the limits are: - \(\lim_{x \to 0^-} f(x) = -\infty\) - \(\lim_{x \to -\infty} f(x) = +\infty\)

Based on the keypoints we derived in the previous step, we can now sketch the graph of the function \(f(x)=\ln |x|\). The graph will have a vertical asymptote at \(x=0\). To the right of the vertical asymptote, the graph will resemble the natural logarithm function, and to the left, it will resemble a mirror image of the natural logarithm function reflected over the y-axis.

Now we will explain how the graph shows that the derivative of the function \(f^{\prime}(x)=\frac{1}{x}\). Let's consider the points \((x_0, y_0)\) on the graph, where \(y_0=\ln|x_0|\). The slope of the tangent line to the graph of \(f(x)\) at \((x_0, y_0)\) represents the derivative of \(f(x)\) at that point, which is \(f^{\prime}(x_0)\). Looking at the graph, we see that the slope becomes flatter as we move away from the vertical asymptote along the x-axis. This means the slope (derivative) is decreasing as x increases in magnitude, suggesting a reciprocal relationship between the slope and x. So, based on the graph, we can understand that \(f^{\prime}(x)\) is of the form \(\frac{1}{x}\), which mathematically confirms the given derivative.