Math Problem Statement

Determine the intervals on which the following function is concave up or concave down. Identify any inflection points. f left parenthesis x right parenthesisequalsStartRoot 4 x EndRoot ln left parenthesis x right parenthesis Question content area bottom Part 1 Determine the intervals on which the following functions are concave up or concave down. Select the correct choice​ and, if​ necessary, fill in the answer​ box(es) to complete your choice. A. The function is concave up on    enter your response here and concave down on    enter your response here. ​(Simplify your answers. Type your answers in interval notation. Use a comma to separate answers as needed. Use integers or fractions for any numbers in the​ expression.) B. The function is concave down on    enter your response here and the function is never concave up. ​(Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed. Use integers or fractions for any numbers in the​ expression.) C. The function is concave up on    enter your response here and the function is never concave down. ​(Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed. Use integers or fractions for any numbers in the​ expression.) D. The function is never concave up nor concave down. Part 2 Locate any inflection points of f. Select the correct choice​ and, if​ necessary, fill in the answer box to complete your choice. A. An inflection point occurs at xequals    enter your response here. ​(Type an integer or a simplified fraction. Use a comma to separate answers as​ needed.) B. There are no inflection points for f.

Solution

To determine the concavity of the function f(x)=4xln(x)f(x) = \sqrt{4x} \ln(x), we need to calculate the second derivative and analyze its sign.

Step 1: Find the first derivative f(x)f'(x)

Let's rewrite f(x)f(x) to make differentiation simpler: f(x)=4xln(x)=2xln(x)=2x1/2ln(x)f(x) = \sqrt{4x} \ln(x) = 2\sqrt{x} \ln(x) = 2x^{1/2} \ln(x)

Using the product rule for differentiation, where u=2x1/2u = 2x^{1/2} and v=ln(x)v = \ln(x):

  1. u=1xu' = \frac{1}{\sqrt{x}}
  2. v=1xv' = \frac{1}{x}

Then, f(x)=uv+uv=(1x)ln(x)+2x1/21x=ln(x)x+1x=ln(x)+1xf'(x) = u'v + uv' = \left( \frac{1}{\sqrt{x}} \right) \ln(x) + 2x^{1/2} \cdot \frac{1}{x} = \frac{\ln(x)}{\sqrt{x}} + \frac{1}{\sqrt{x}} = \frac{\ln(x) + 1}{\sqrt{x}}

Step 2: Find the second derivative f(x)f''(x)

Now, differentiate f(x)=ln(x)+1xf'(x) = \frac{\ln(x) + 1}{\sqrt{x}} again.

Using the quotient rule for differentiation, where:

  • Numerator: g(x)=ln(x)+1g(x) = \ln(x) + 1, so g(x)=1xg'(x) = \frac{1}{x}
  • Denominator: h(x)=x=x1/2h(x) = \sqrt{x} = x^{1/2}, so h(x)=12xh'(x) = \frac{1}{2\sqrt{x}}

The second derivative f(x)f''(x) is given by: f(x)=ghghh2=(1x)x(ln(x)+1)12xxf''(x) = \frac{g' h - g h'}{h^2} = \frac{\left(\frac{1}{x}\right) \sqrt{x} - (\ln(x) + 1) \cdot \frac{1}{2\sqrt{x}}}{x} Simplifying each term will show where f(x)f''(x) is positive (concave up) or negative (concave down).

Step 3: Determine concavity and inflection points

  1. Solve f(x)=0f''(x) = 0 to find possible inflection points.
  2. Analyze the intervals based on where f(x)f''(x) changes sign.

Once calculated, we can determine whether to select:

  • Option A if the function is both concave up and concave down on different intervals.
  • Option B or C if the function is only concave up or concave down across all intervals.
  • Option D if the function has no concavity.

Would you like further assistance with detailed interval calculations or the specific selection of intervals for concavity? Here are related questions you might find helpful:

  1. What is the process of finding intervals of concavity for any given function?
  2. How does the presence of a logarithmic term affect concavity?
  3. How do inflection points relate to the behavior of the first derivative?
  4. What are typical points where functions with square roots or logarithms change concavity?
  5. How does interval notation help in expressing concavity?

Tip: When working with functions involving ln(x)\ln(x), ensure the domain x>0x > 0 is respected for valid solutions.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Concavity
Inflection Points
Product Rule
Quotient Rule
Logarithmic Functions
Square Root Functions

Formulas

First derivative: f'(x) = (ln(x) + 1)/√x
Second derivative: f''(x) = (1/x)√x - (ln(x) + 1)/(2√x)

Theorems

Product Rule of Differentiation
Quotient Rule of Differentiation

Suitable Grade Level

Grades 11-12