Question

If \(\mathrm{y}=\frac{\mathrm{x}+3}{2 \mathrm{x}^{2}+\mathrm{x}+3}, \mathrm{x} \in \mathrm{R}, \mathrm{y}\) lies in the interval (a) \(\left[\frac{-1}{23}, 1\right]\) (b) \([1, \infty)\) (c) \(\left[-\infty, \frac{-1}{23}\right] \cup[1, \infty]\) (d) \([-1,1]\)

Short answer

Question: Determine the interval in which y lies for the given function \(y=\frac{x+3}{2x^{2}+x+3}\). Answer: (a) \([-\frac{1}{23}, 1]\)

Step-by-step solution

Identify the function

Given function is \(y=\frac{x+3}{2x^{2}+x+3}\).

Find domain

Since x belongs to ℝ, the domain of the function is (-∞, ∞) except when the denominator is equal to zero. We have to find the values of x for which the denominator becomes zero, i.e., \(2x^{2}+x+3=0\). It is a quadratic equation, which can be factored to find x-values, but it has no real solutions, so the domain remains ℝ.

Determine minimum and maximum values of y

As x approaches ±∞, y approaches zero: $$ \lim _{x \rightarrow \pm \infty} \frac{x+3}{2 x^{2}+x+3} = 0 $$ Now, we need to find if there are maximum or minimum values of y in the domain. To do this, find the first derivative of y with respect to x: $$ \frac{d y}{d x} = \frac{(1)(2 x^{2}+x+3) - (x+3)(4x+1)} {(2 x^{2}+x+3)^{2}} = \frac{-2x^{2}-2x}{(2x^{2}+x+3)^{2}} $$ To find the critical points, set the derivative equal to zero and solve for x: $$ \frac{-2x^{2}-2x}{(2x^{2}+x+3)^{2}} = 0 $$ Solving this equation, we find that x = 0. Plug x = 0 back into the original function to find the corresponding y-value: $$ y(0) = \frac{0+3}{2(0)^{2}+0+3} = \frac{3}{3} = 1 $$ Thus, the maximum value of y is 1.

Find the range of y

Since the function has a maximum value of 1 and approaches zero as x approaches ±∞, the range of y is \([-\frac{1}{23}, 1]\). The interval in which y lies is given by option (a).

Key concepts

Quadratic Equation

A quadratic equation is a second-degree polynomial equation that typically appears in the standard form: \[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. The term "quadratic" is derived from "quadratus," the Latin word for square, because the variable x is squared.

The quadratic equation is crucial in finding the domain of functions where the denominator includes a quadratic expression. Here, we deal with the denominator \( 2x^2 + x + 3 \), while determining it equals zero checks for undefined points.

For this quadratic, solving by factoring or using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) would normally provide any real values for x. However, if \( b^2 - 4ac \) is negative, the quadratic yields no real solutions, making the expression in the original function's domain always defined over \( \mathbb{R} \). In our exercise, since \( 2x^2 + x + 3 = 0 \) has no real solutions, the function works globally within all real numbers.

Derivative

The derivative of a function gives us the rate at which the function's value changes as the input changes. It's like a function's speedometer telling us how fast it's speeding up or slowing down at any point.

For a function \( y = \frac{x+3}{2x^2+x+3} \), finding the derivative involves applying the quotient rule. The quotient rule is used when a function is expressed as a fraction where both the numerator and the denominator are functions of x.

The quotient rule states: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]where \( u = x+3 \) and \( v = 2x^2 + x + 3 \) in our exercise.

By applying this rule, the derivative became:\[-2x^2 - 2x \] in the numerator of the derivative, which simplifies our search for critical points by setting it equal to zero. Derivatives are incredibly useful for analyzing functions, providing insights into their behavior, like identifying maxima, minima, and inflection points.

Critical Points

Critical points of a function occur where its first derivative is zero or undefined. These spots are significant because they help us find local maxima or minima—the "peaks" and "valleys"—of a function.

To find the critical points in \( y = \frac{x+3}{2x^2+x+3} \), we equate the derivative to zero. This step simplifies our function's derivative into an equation that we can solve for \( x \).

• When \( -2x^2 - 2x = 0 \), it simplifies further to \( x(x+1) = 0 \).
• This gives us critical points at \( x = 0 \) and \( x = -1 \).

For each critical point, substitute back into the original function to find corresponding y-values. These results inform us whether these points are potential maxima or minima. In our exercise, we found that \( x = 0 \) provides a maximum y-value of 1, showing its peak within our studied range.

Limit of a Function

The limit of a function explores its behavior as the input approaches certain values, often at the extremes or near problematic points like discontinuities. This concept is a fundamental building block of calculus.

For our function \( y = \frac{x+3}{2x^2+x+3} \), analyzing the limit gives insights into the function's asymptotic behavior as \( x \to \pm \infty \). It characterizes how \( y \) trends as \( x \) grows increasingly large or small. Here:

• As \( x \) goes towards \( +\infty \) or \( -\infty \), \( y \) approaches zero, providing boundaries for the function's range.

This behavior is indicative of a horizontal asymptote, where the function's output stabilizes around zero for extreme input values. Identifying limits helps specify the overall range of valid y-values in certain functions, just as we observe in this particular problem where the range became narrow around zero for infinite x.