The given function is:

$f(x) = \sqrt{2x^2 + 5x + 2}$

Let’s break down a few key concepts about this function.

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1. Domain of the Function

Since $f(x)$ contains a square root, the expression inside the square root $2x^2 + 5x + 2$ must be non-negative:

$2x^2 + 5x + 2 \geq 0$

To determine the domain, we solve the quadratic inequality.

1. Find the roots:
   Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = 5$, and $c = 2$.

   = \frac{-5 \pm \sqrt{9}}{4} = \frac{-5 \pm 3}{4}$$ So the roots are: $$x = -\frac{1}{2}, \quad x = -2$$

2. Solve the inequality $2x^2 + 5x + 2 \geq 0$:
   We need to determine the sign of the quadratic expression in the intervals formed by these roots:

   • $(-\infty, -2)$
   • $(-2, -\frac{1}{2})$
   • $(-\frac{1}{2}, \infty)$

   For a quadratic of the form $ax^2 + bx + c$, if $a > 0$, the expression is positive outside the roots. Therefore, the solution to the inequality is:

   $(-\infty, -2] \cup \left[-\frac{1}{2}, \infty\right)$

   Domain:
   The domain of the function is: $(-\infty, -2] \cup \left[-\frac{1}{2}, \infty\right)$

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2. Range of the Function

The function involves a square root, so the range will be non-negative, i.e., $f(x) \geq 0$. The minimum value of the square root occurs when the expression inside is minimized. This is tricky because it depends on $2x^2 + 5x + 2$, which is non-negative across the domain.

Therefore, the range is:

$[0, \infty)$

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3. Derivative (Slope of the Function)

To find the rate of change or slope of the function, we need the derivative using the chain rule.

$f'(x) = \frac{1}{2\sqrt{2x^2 + 5x + 2}} \cdot \frac{d}{dx}(2x^2 + 5x + 2)$

$f'(x) = \frac{1}{2\sqrt{2x^2 + 5x + 2}} \cdot (4x + 5)$

Thus, the derivative is:

$f'(x) = \frac{4x + 5}{2\sqrt{2x^2 + 5x + 2}}$

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4. Behavior of the Function (Increasing/Decreasing)

• The function is increasing when $f'(x) > 0$.
• This happens when $4x + 5 > 0$, i.e., $x > -\frac{5}{4}$.
• The function is decreasing when $x < -\frac{5}{4}$.

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5. Asymptotic Behavior

There are no vertical asymptotes since the function is defined in terms of a square root polynomial. Also, as $x \to \infty$ or $x \to -\infty$, the function will grow without bound, meaning:

$\lim_{x \to \pm \infty} f(x) = \infty$

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Would you like more details or have any questions?

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Related Questions

1. How do you solve quadratic inequalities step by step?
2. What are the rules for applying the chain rule in differentiation?
3. How do we determine where a function is increasing or decreasing?
4. How do we find the range of complex functions involving square roots?
5. What is the meaning of asymptotic behavior for polynomial-root functions?

Tip: When solving quadratic inequalities, always test values within each interval to confirm the sign.