Discover how IDNLearn.com can help you find the answers you need quickly and easily. Join our knowledgeable community and get detailed, reliable answers to all your questions.
Sagot :
Alright, let's break down the given problem and solve it step-by-step to find the minimum integer value in the range of the function [tex]\( f(x) \)[/tex].
### Step 1: Understanding the Function
The given function is:
[tex]\[ f(x) = x^2 - 4x + \left( \frac{a^2 + b}{b + c} + \frac{b^2 + c}{a + c} + \frac{c^2 + a}{b + a} \right) \][/tex]
### Step 2: Simplifying the Constant Term
The constant term in the function is:
[tex]\[ \frac{a^2 + b}{b + c} + \frac{b^2 + c}{a + c} + \frac{c^2 + a}{b + a} \][/tex]
Given that [tex]\( a + b + c = 1 \)[/tex], we need to work on simplifying this term.
### Step 3: Finding the Minimum of the Quadratic Expression
The quadratic expression [tex]\( x^2 - 4x + C \)[/tex] is a standard quadratic function where [tex]\(C\)[/tex] is a constant derived from the problem. The minimum value of this quadratic expression can be found using the vertex formula for [tex]\(ax^2 + bx + c\)[/tex], where the vertex [tex]\(x\)[/tex] coordinate is given by:
[tex]\[ x = \frac{-b}{2a} \][/tex]
For our function:
[tex]\[ a = 1, \; b = -4 \][/tex]
So,
[tex]\[ x = \frac{-(-4)}{2 \cdot 1} = \frac{4}{2} = 2 \][/tex]
### Step 4: Evaluating the Function at the Vertex
Now, substituting [tex]\( x = 2 \)[/tex] back into the function:
[tex]\[ f(2) = 2^2 - 4 \cdot 2 + C \][/tex]
[tex]\[ f(2) = 4 - 8 + C \][/tex]
[tex]\[ f(2) = -4 + C \][/tex]
### Step 5: Finding the Constant Term
Now we need to determine the value of the constant term [tex]\( C \)[/tex]:
[tex]\[ C = \frac{a^2 + b}{b + c} + \frac{b^2 + c}{a + c} + \frac{c^2 + a}{b + a} \][/tex]
Let's use symmetry and the condition [tex]\( a + b + c = 1 \)[/tex] to simplify further. By trial of typical symmetrical values:
Let's assume [tex]\( a = b = c \)[/tex]
Since [tex]\( a + b + c = 1 \)[/tex], if [tex]\( a = b = c \)[/tex], then:
[tex]\[ 3a = 1 \Rightarrow a = \frac{1}{3} \][/tex]
Thus:
[tex]\[ C = \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} + \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} + \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} \][/tex]
Simplify each term individually:
[tex]\[ \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{2}{3}} = \frac{\frac{1}{9} + \frac{1}{3}}{\frac{2}{3}} \][/tex]
[tex]\[ \frac{\frac{1}{9} + \frac{3}{9}}{\frac{2}{3}} = \frac{\frac{4}{9}}{\frac{2}{3}} = \frac{4}{9} \cdot \frac{3}{2} = \frac{4}{3 \cdot 3} \cdot 3 = \frac{2}{3} \][/tex]
Adding up the three terms:
[tex]\[ C = \frac{2}{3} + \frac{2}{3} + \frac{2}{3} = 2 \][/tex]
### Step 6: Evaluate the Minimum Value of the Function
Using [tex]\( C = 2 \)[/tex]:
[tex]\[ f(2) = -4 + 2 = -2 \][/tex]
### Conclusion: Minimum Integer Value
The minimum value of the function [tex]\( f(x) \)[/tex] is [tex]\(-2\)[/tex], which is already an integer.
Thus, the minimum integer value in the range of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{-2} \][/tex]
### Step 1: Understanding the Function
The given function is:
[tex]\[ f(x) = x^2 - 4x + \left( \frac{a^2 + b}{b + c} + \frac{b^2 + c}{a + c} + \frac{c^2 + a}{b + a} \right) \][/tex]
### Step 2: Simplifying the Constant Term
The constant term in the function is:
[tex]\[ \frac{a^2 + b}{b + c} + \frac{b^2 + c}{a + c} + \frac{c^2 + a}{b + a} \][/tex]
Given that [tex]\( a + b + c = 1 \)[/tex], we need to work on simplifying this term.
### Step 3: Finding the Minimum of the Quadratic Expression
The quadratic expression [tex]\( x^2 - 4x + C \)[/tex] is a standard quadratic function where [tex]\(C\)[/tex] is a constant derived from the problem. The minimum value of this quadratic expression can be found using the vertex formula for [tex]\(ax^2 + bx + c\)[/tex], where the vertex [tex]\(x\)[/tex] coordinate is given by:
[tex]\[ x = \frac{-b}{2a} \][/tex]
For our function:
[tex]\[ a = 1, \; b = -4 \][/tex]
So,
[tex]\[ x = \frac{-(-4)}{2 \cdot 1} = \frac{4}{2} = 2 \][/tex]
### Step 4: Evaluating the Function at the Vertex
Now, substituting [tex]\( x = 2 \)[/tex] back into the function:
[tex]\[ f(2) = 2^2 - 4 \cdot 2 + C \][/tex]
[tex]\[ f(2) = 4 - 8 + C \][/tex]
[tex]\[ f(2) = -4 + C \][/tex]
### Step 5: Finding the Constant Term
Now we need to determine the value of the constant term [tex]\( C \)[/tex]:
[tex]\[ C = \frac{a^2 + b}{b + c} + \frac{b^2 + c}{a + c} + \frac{c^2 + a}{b + a} \][/tex]
Let's use symmetry and the condition [tex]\( a + b + c = 1 \)[/tex] to simplify further. By trial of typical symmetrical values:
Let's assume [tex]\( a = b = c \)[/tex]
Since [tex]\( a + b + c = 1 \)[/tex], if [tex]\( a = b = c \)[/tex], then:
[tex]\[ 3a = 1 \Rightarrow a = \frac{1}{3} \][/tex]
Thus:
[tex]\[ C = \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} + \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} + \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} \][/tex]
Simplify each term individually:
[tex]\[ \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{2}{3}} = \frac{\frac{1}{9} + \frac{1}{3}}{\frac{2}{3}} \][/tex]
[tex]\[ \frac{\frac{1}{9} + \frac{3}{9}}{\frac{2}{3}} = \frac{\frac{4}{9}}{\frac{2}{3}} = \frac{4}{9} \cdot \frac{3}{2} = \frac{4}{3 \cdot 3} \cdot 3 = \frac{2}{3} \][/tex]
Adding up the three terms:
[tex]\[ C = \frac{2}{3} + \frac{2}{3} + \frac{2}{3} = 2 \][/tex]
### Step 6: Evaluate the Minimum Value of the Function
Using [tex]\( C = 2 \)[/tex]:
[tex]\[ f(2) = -4 + 2 = -2 \][/tex]
### Conclusion: Minimum Integer Value
The minimum value of the function [tex]\( f(x) \)[/tex] is [tex]\(-2\)[/tex], which is already an integer.
Thus, the minimum integer value in the range of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{-2} \][/tex]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.