Use the first derivative test to determine the location of each local extremum and the value of the function at this extremum. \( f(x)=x \ln |x| \)

Alright, I need to determine the local extrema of the function \( f(x) = x \ln |x| \) using the first derivative test. Let's start by understanding the function and its domain. First, the function \( f(x) = x \ln |x| \) is defined for all real numbers except \( x = 0 \), where the logarithm is undefined. So, the domain is \( x \in (-\infty, 0) \cup (0, \infty) \). Next, I'll find the first derivative of \( f(x) \) to identify critical points. To differentiate \( f(x) \), I'll use the product rule since it's a product of two functions: \( x \) and \( \ln |x| \). The product rule states that if \( f(x) = u(x) \cdot v(x) \), then \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Here, \( u(x) = x \) and \( v(x) = \ln |x| \). First, find \( u'(x) \): \( u'(x) = 1 \) Next, find \( v'(x) \): \( v(x) = \ln |x| \) Since \( \ln |x| \) is the natural logarithm of the absolute value of \( x \), its derivative is: \( v'(x) = \frac{1}{x} \) for \( x > 0 \) and \( v'(x) = \frac{1}{x} \) for \( x < 0 \) because the derivative of \( \ln |x| \) is the same in both intervals. Now, apply the product rule: \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \) \( f'(x) = 1 \cdot \ln |x| + x \cdot \frac{1}{x} \) Simplify the expression: \( f'(x) = \ln |x| + 1 \) Now, set the first derivative equal to zero to find critical points: \( \ln |x| + 1 = 0 \) \( \ln |x| = -1 \) Exponentiate both sides to solve for \( |x| \): \( |x| = e^{-1} \) \( |x| = \frac{1}{e} \) This gives two critical points: \( x = \frac{1}{e} \) and \( x = -\frac{1}{e} \) Now, I'll analyze the sign of the first derivative around these critical points to determine the nature of the extrema. 1. **For \( x = \frac{1}{e} \):** - Choose a test point \( x = 0.5 \) (which is greater than \( \frac{1}{e} \approx 0.3679 \)): \( f'(0.5) = \ln |0.5| + 1 = \ln 0.5 + 1 \approx -0.6931 + 1 = 0.3069 > 0 \) - Choose a test point \( x = 0.3 \) (which is less than \( \frac{1}{e} \)): \( f'(0.3) = \ln |0.3| + 1 = \ln 0.3 + 1 \approx -1.204 + 1 = -0.204 < 0 \) - Since the derivative changes from negative to positive at \( x = \frac{1}{e} \), there is a local minimum at this point. 2. **For \( x = -\frac{1}{e} \):** - Choose a test point \( x = -0.5 \) (which is less than \( -\frac{1}{e} \)): \( f'(-0.5) = \ln |-0.5| + 1 = \ln 0.5 + 1 \approx -0.6931 + 1 = 0.3069 > 0 \) - Choose a test point \( x = -0.3 \) (which is greater than \( -\frac{1}{e} \)): \( f'(-0.3) = \ln |-0.3| + 1 = \ln 0.3 + 1 \approx -1.204 + 1 = -0.204 < 0 \) - Since the derivative changes from positive to negative at \( x = -\frac{1}{e} \), there is a local maximum at this point. Now, I'll calculate the function values at these critical points to find the extrema. 1. **At \( x = \frac{1}{e} \):** \( f\left(\frac{1}{e}\right) = \frac{1}{e} \ln \left|\frac{1}{e}\right| = \frac{1}{e} \cdot (-1) = -\frac{1}{e} \) 2. **At \( x = -\frac{1}{e} \):** \( f\left(-\frac{1}{e}\right) = -\frac{1}{e} \ln \left|-\frac{1}{e}\right| = -\frac{1}{e} \cdot (-1) = \frac{1}{e} \) **Summary of Findings:** - **Local Minimum:** At \( x = \frac{1}{e} \), the function value is \( -\frac{1}{e} \). - **Local Maximum:** At \( x = -\frac{1}{e} \), the function value is \( \frac{1}{e} \).