Math Problem Statement

The given function is f(x) = (a+1)x^2 - (a+3)x + a - 2. Solve the following problems: (I) If the graph of f(x) is symmetric about the line x = 1, find the value of a. (II) If a = -1, solve the inequality f(x) <= (-2x^2 + 3x + 7) / (x + 1). (III) If for any x ∈ (0, +∞), f(x) >= x^2 - 2x - 2 always holds, find the range of a.

Solution

The image contains three mathematical problems based on the function:

f(x)=(a+1)x2(a+3)x+a2f(x) = (a+1)x^2 - (a+3)x + a - 2

Let's break down and solve each of these parts:


(I) The problem asks: If the graph of f(x)f(x) is symmetric with respect to the line x=1x = 1, find the value of aa.

  • A quadratic function is symmetric about a vertical line x=px = p when its vertex lies on that line. The vertex of a quadratic function f(x)=Ax2+Bx+Cf(x) = Ax^2 + Bx + C is located at x=B2Ax = -\frac{B}{2A}.
  • In our case:
    • A=(a+1)A = (a+1)
    • B=(a+3)B = -(a+3)
    • The vertex must lie at x=1x = 1.

Therefore:

(a+3)2(a+1)=1-\frac{-(a+3)}{2(a+1)} = 1

Simplifying:

a+32(a+1)=1\frac{a+3}{2(a+1)} = 1

Multiply both sides by 2(a+1)2(a+1):

a+3=2(a+1)a + 3 = 2(a + 1)

Expanding the right-hand side:

a+3=2a+2a + 3 = 2a + 2

Solving for aa:

a=1a = 1

Thus, the value of aa is 1.


(II) Given a=1a = -1, solve the inequality:

f(x)2x2+3x+7x+1f(x) \leq \frac{-2x^2 + 3x + 7}{x+1}

Substituting a=1a = -1 into the expression for f(x)f(x):

f(x)=(0)x2(2)x+(3)=2x3f(x) = (0)x^2 - (-2)x + (-3) = 2x - 3

So, we need to solve:

2x32x2+3x+7x+12x - 3 \leq \frac{-2x^2 + 3x + 7}{x+1}

First, simplify the right-hand side:

2x2+3x+7x+1\frac{-2x^2 + 3x + 7}{x+1}

This can be simplified by dividing the numerator by x+1x + 1 using polynomial long division, yielding:

2x+5+2x+1-2x + 5 + \frac{2}{x+1}

Thus, the inequality becomes:

2x32x+5+2x+12x - 3 \leq -2x + 5 + \frac{2}{x+1}

Now, solve this inequality.


(III) Solve for the values of aa such that for any x(0,)x \in (0, \infty), the inequality f(x)x22x2f(x) \geq x^2 - 2x - 2 holds.

Substitute f(x)=(a+1)x2(a+3)x+a2f(x) = (a+1)x^2 - (a+3)x + a - 2 into the inequality:

(a+1)x2(a+3)x+(a2)x22x2(a+1)x^2 - (a+3)x + (a-2) \geq x^2 - 2x - 2

Rearranging:

(a+1)x2(a+3)x+(a2)x2+2x+20(a+1)x^2 - (a+3)x + (a-2) - x^2 + 2x + 2 \geq 0

Simplifying:

(a+11)x2+(a3+2)x+(a2+2)0(a+1 - 1)x^2 + (-a - 3 + 2)x + (a - 2 + 2) \geq 0

ax2(a+1)x+a0a x^2 - (a+1)x + a \geq 0

This quadratic inequality needs to be solved to find the values of aa such that the inequality holds for x(0,)x \in (0, \infty).


Would you like further clarification or detailed solutions for any part? Here are 5 related questions to explore this topic further:

  1. How do you apply the vertex formula to find symmetry lines in quadratic functions?
  2. What is the process for solving quadratic inequalities?
  3. How do you perform polynomial long division?
  4. What are the conditions for a quadratic function to be positive over an interval?
  5. How does the value of aa affect the shape and position of the parabola defined by f(x)f(x)?

Tip: When solving inequalities involving rational functions, always check for undefined points where the denominator equals zero.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Symmetry
Inequalities
Polynomial Division

Formulas

Quadratic function: f(x) = ax^2 + bx + c
Vertex formula: x = -b / (2a)
Polynomial division

Theorems

Properties of quadratic functions
Solving quadratic inequalities

Suitable Grade Level

Grades 10-12