Math Problem Statement

Find the open intervals on which the function is increasing and those on which it is decreasing for f(x) = x * ln(x). Identify the function's local extreme values, if any, and where they occur.

Solution

To solve this problem, let’s go through each part carefully.

Given function:

f(x)=xln(x)f(x) = x \ln(x)

Part (a): Find the intervals on which the function is increasing and decreasing.

  1. Find the derivative f(x)f'(x):

    Using the product rule for differentiation where f(x)=xln(x)f(x) = x \ln(x): f(x)=ddx(x)ln(x)+xddx(ln(x))f'(x) = \frac{d}{dx}(x) \cdot \ln(x) + x \cdot \frac{d}{dx}(\ln(x)) Simplifying each term: f(x)=1ln(x)+x1xf'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} f(x)=ln(x)+1f'(x) = \ln(x) + 1

  2. Set f(x)=0f'(x) = 0 to find critical points:

    ln(x)+1=0\ln(x) + 1 = 0 ln(x)=1\ln(x) = -1 Solving for xx: x=e1=1ex = e^{-1} = \frac{1}{e}

    So, we have a critical point at x=1ex = \frac{1}{e}.

  3. Determine the sign of f(x)f'(x) on the intervals around the critical point x=1ex = \frac{1}{e}:

    • For x<1ex < \frac{1}{e}: Choose a point like x=0.1x = 0.1.

      • f(0.1)=ln(0.1)+1f'(0.1) = \ln(0.1) + 1, and since ln(0.1)<1\ln(0.1) < -1, f(0.1)<0f'(0.1) < 0, so f(x)f(x) is decreasing on this interval.
    • For x>1ex > \frac{1}{e}: Choose a point like x=1x = 1.

      • f(1)=ln(1)+1=0+1=1f'(1) = \ln(1) + 1 = 0 + 1 = 1, so f(1)>0f'(1) > 0, meaning f(x)f(x) is increasing on this interval.

    Therefore, the function f(x)f(x) is increasing on the interval (1e,)\left(\frac{1}{e}, \infty\right).

Answer for Part (a):

The correct answer is A: The function ff is increasing on the open interval (1e,)\left(\frac{1}{e}, \infty\right).

Part (b): Identify local extreme values, if any.

Since f(x)f(x) changes from decreasing to increasing at x=1ex = \frac{1}{e}, there is a local minimum at this point.

  • Local Minimum Value: Substitute x=1ex = \frac{1}{e} into f(x)f(x): f(1e)=1eln(1e)f\left(\frac{1}{e}\right) = \frac{1}{e} \ln\left(\frac{1}{e}\right) =1e(1)=1e= \frac{1}{e} \cdot (-1) = -\frac{1}{e}

Therefore, there is a local minimum of 1e-\frac{1}{e} at x=1ex = \frac{1}{e}.


Summary:

  • Intervals of Increase: (1e,)\left(\frac{1}{e}, \infty\right)
  • Local Minimum: 1e-\frac{1}{e} at x=1ex = \frac{1}{e}

Do you need further clarification on any step?


Here are some related questions for deeper understanding:

  1. How would you find intervals of concavity for the function f(x)=xln(x)f(x) = x \ln(x)?
  2. Can you determine if the local minimum at x=1ex = \frac{1}{e} is also a global minimum?
  3. How would the behavior of f(x)=xln(x)f(x) = x \ln(x) change if we considered negative values of xx?
  4. What happens to the function f(x)=xln(x)f(x) = x \ln(x) as xx approaches 0 from the right?
  5. How would you calculate the derivative of a similar function, such as g(x)=x2ln(x)g(x) = x^2 \ln(x)?

Tip: Always check the domain of the function before starting any calculus operations to ensure your intervals are valid.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Critical Points
Intervals of Increase and Decrease

Formulas

f(x) = x * ln(x)
Derivative: f'(x) = ln(x) + 1

Theorems

Product Rule
Critical Points in Calculus

Suitable Grade Level

College Level Calculus