Let g(x) = (1 + α^2)x^2 + 3αx + 9/4 and if m(α) be the minimum value of g(x) as α varies, then the range of m(α) is?
Understand the Problem
The question asks for the range of the minimum value of the function g(x) defined in the problem as α varies. It requires determining the correct interval from the provided options based on the behaviors of the function.
Answer
The range of $m(\alpha)$ is $[0, \frac{9}{4})$.
Answer for screen readers
The range of $m(\alpha)$ is $[0, \frac{9}{4})$.
Steps to Solve
- Identify the function's form
The function given is
$$ g(x) = (1 + \alpha^2)x^2 + 3\alpha x + \frac{9}{4}. $$
This is a quadratic function in the variable $x$.
- Find the minimum value of the quadratic function
For a quadratic function $ax^2 + bx + c$, the minimum value (if $a > 0$) occurs at
$$ x = -\frac{b}{2a}. $$
Here, $a = 1 + \alpha^2$ and $b = 3\alpha$. Thus,
$$ x_{min} = -\frac{3\alpha}{2(1 + \alpha^2)}. $$
- Calculate the minimum value $m(\alpha)$ of $g(x)$
Substituting $x_{min}$ back into $g(x)$, we find
$$ m(\alpha) = g\left(-\frac{3\alpha}{2(1 + \alpha^2)}\right). $$
Simplifying this requires substituting the expression for $x_{min}$ into the original function $g(x)$.
- Determine the range of $m(\alpha)$
After simplifying $m(\alpha)$, we analyze its values as $\alpha$ varies. The goal is to identify the range of possible outputs.
- Identify the range from the provided options
The options given for the range of $m(\alpha)$ are:
- A. $[0, \frac{9}{4})$
- B. $(0, \frac{9}{4}]$
- C. $(0, \frac{4}{9}]$
- D. $[0, \frac{4}{9}]$
Analyze the behavior of the function to select the correct interval.
The range of $m(\alpha)$ is $[0, \frac{9}{4})$.
More Information
The quadratic function reaches its minimum at a specific $x$ value that depends on $\alpha$. As $\alpha$ varies, the minimum value can approach 0 and will be less than $\frac{9}{4}$.
Tips
- Not recognizing that the function is quadratic and assuming it behaves linearly.
- Forgetting to check the sign of the coefficient $a$ to confirm it's a minimum.
- Failing to substitute the correct $x_{min}$ into the function while calculating $m(\alpha)$.
AI-generated content may contain errors. Please verify critical information
