Let y=(x^(2)+4x+5)/(x+2), where x!=-2. Prove that if x inR then y cannot take values between -2 and 2 .

Question

Let  y=(x^(2)+4x+5)/(x+2), where  x!=-2. Prove that if  x inR then  y cannot take values between -2 and 2 .

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Simplify Expression for y: Let's start by simplifying the expression for y by performing polynomial division or factoring, if possible. y = (x^2 + 4x + 5) / (x + 2) We can try to factor the numerator, but since the roots of x^2 + 4x + 5 are complex, we cannot factor it over the real numbers. Therefore, we will perform polynomial division.Perform Polynomial Division: Perform polynomial division of the numerator by the denominator. (x^2 + 4x + 5) ÷ (x + 2) gives us x + 2 with a remainder of 1. So, y = x + 2 + 1/(x + 2)Analyze Expression y: Now, let's analyze the expression y = x + 2 + 1/(x + 2). We know that x cannot be -2, so the term 1/(x + 2) is always defined. For y to be between -2 and 2, the following inequality must be true: -2 < x + 2 + 1/(x + 2) < 2Subtract 2 from Inequality: Subtract 2 from all parts of the inequality to isolate the fraction on one side. -4 < x + 1/(x + 2) < 0Consider Left Part of Inequality: Now, let's consider the two parts of the inequality separately. First, let's look at the left part: -4 < x + 1/(x + 2). For x > -2, the term 1/(x + 2) is positive, and thus x + 1/(x + 2) is always greater than x. Since x can be any real number greater than -2, there will always be values of x for which x + 1/(x + 2) is greater than -4.Consider Right Part of Inequality: Now, let's look at the right part of the inequality: x + 1/(x + 2) < 0. For x > -2, the term 1/(x + 2) is positive, and thus x + 1/(x + 2) is always greater than x. For x + 1/(x + 2) to be less than 0, x would have to be negative and its absolute value would have to be greater than 1/(x + 2). However, as x approaches -2 from the right, 1/(x + 2) becomes arbitrarily large, so there is no real number x for which x + 1/(x + 2) is less than 0.Conclusion: Since there is no real number x for which x + 1/(x + 2) is less than 0, the original inequality -2 < x + 2 + 1/(x + 2) < 2 cannot be satisfied for any real x. Therefore, y cannot take values between -2 and 2 for any real x.

Let $y=\frac{x^{2}+4 x+5}{x+2}$ , where $x \neq-2$ . $\newline$ Prove that if $x \in \mathbb{R}$ then $y$ cannot take values between $-2$ and $2$ .

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Let y=x2+4x+5x+2 y=\frac{x^{2}+4 x+5}{x+2} y=x+2x2+4x+5​, where x≠−2 x \neq-2 x=−2.\newlineProve that if x∈R x \in \mathbb{R} x∈R then y y y cannot take values between −2-2−2 and 222 .

Let $y=\frac{x^{2}+4 x+5}{x+2}$ , where $x \neq-2$ . $\newline$ Prove that if $x \in \mathbb{R}$ then $y$ cannot take values between $-2$ and $2$ .