Math Problem Statement

A function f is defined by f(x) = |x|^m |x - 1|^n for x ∈ R. The local maximum value of the function is, (m, n ∈ N), (a) 1 (b) m^n n^m / m^m n^n (c) (mn)^(mn) / (m+n)^(m+n) (d) m^m n^n / (m+n)^(m+n)

Solution

The given problem involves determining the local maximum of the function:
f(x)=xmx1n,xR,m,nN.f(x) = |x|^m |x - 1|^n, \quad x \in \mathbb{R}, \, m, n \in \mathbb{N}.

To find the local maximum value, we proceed as follows:


Step 1: Analyze the function

The function f(x)f(x) is the product of two absolute values, raised to the powers mm and nn:

  1. xm|x|^m: This term is zero at x=0x = 0, increases as x|x| increases, and is symmetric about x=0x = 0.
  2. x1n|x - 1|^n: This term is zero at x=1x = 1, increases as x1|x - 1| moves away from 11, and is asymmetric.

The combined behavior creates a local maximum at some point between 00 and 11, where the product achieves a balance between the two terms.


Step 2: Rewrite f(x)f(x)

Let xx be in [0,1][0, 1]. The absolute values simplify to: f(x)=xm(1x)n,0x1.f(x) = x^m (1 - x)^n, \quad 0 \leq x \leq 1.


Step 3: Find the critical points

To find the critical points, compute the derivative f(x)f'(x): f(x)=ddx(xm(1x)n).f'(x) = \frac{d}{dx} \left( x^m (1 - x)^n \right). Using the product rule: f(x)=mxm1(1x)nnxm(1x)n1.f'(x) = m x^{m-1} (1 - x)^n - n x^m (1 - x)^{n-1}.

Set f(x)=0f'(x) = 0: mxm1(1x)n=nxm(1x)n1.m x^{m-1} (1 - x)^n = n x^m (1 - x)^{n-1}.

Simplify by dividing through by xm1(1x)n1x^{m-1} (1 - x)^{n-1} (valid since x(0,1)x \in (0, 1)): m(1x)=nx.m (1 - x) = n x.

Rearrange to solve for xx: x=mm+n.x = \frac{m}{m + n}.


Step 4: Compute f(x)f(x) at x=mm+nx = \frac{m}{m+n}

Substitute x=mm+nx = \frac{m}{m+n} into f(x)=xm(1x)nf(x) = x^m (1 - x)^n: f(mm+n)=(mm+n)m(1mm+n)n.f \left( \frac{m}{m+n} \right) = \left( \frac{m}{m+n} \right)^m \left( 1 - \frac{m}{m+n} \right)^n.

Simplify 1mm+n1 - \frac{m}{m+n}: 1mm+n=nm+n.1 - \frac{m}{m+n} = \frac{n}{m+n}.

Thus: f(mm+n)=(mm+n)m(nm+n)n.f \left( \frac{m}{m+n} \right) = \left( \frac{m}{m+n} \right)^m \left( \frac{n}{m+n} \right)^n.


Step 5: Final Expression for the Local Maximum

The local maximum value is: fmax=mmnn(m+n)m+n.f_{\text{max}} = \frac{m^m n^n}{(m+n)^{m+n}}.


Step 6: Correct Answer

From the given options, the correct answer is:
(d) (mn)mn(m+n)m+n\frac{(mn)^{mn}}{(m+n)^{m+n}}.


Let me know if you'd like further clarification or detailed steps!


Related Questions:

  1. What is the derivative of xa(1x)bx^a (1-x)^b for general a,b>0a, b > 0?
  2. How do you find the critical points of a product of functions?
  3. How does the symmetry of x|x| and xc|x - c| affect the behavior of functions?
  4. Can you derive the second derivative of f(x)f(x) to confirm the local maximum at x=mm+nx = \frac{m}{m+n}?
  5. How does the value of mm and nn affect the location and value of the maximum?

Tip:

When dealing with products of terms involving absolute values, carefully simplify their behavior in different regions of xx and use symmetry to reduce the complexity of calculations.

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Math Problem Analysis

Mathematical Concepts

Critical points of functions
Maximization of functions
Absolute values and exponents
Derivative and optimization

Formulas

f(x) = x^m (1 - x)^n
f'(x) = m x^(m-1) (1 - x)^n - n x^m (1 - x)^(n-1)
Local maximum value: f_max = (m^m n^n) / (m+n)^(m+n)

Theorems

Critical point theorem for maxima and minima
Power rule for derivatives
Product rule for derivatives

Suitable Grade Level

Grades 11-12