Questions: +1/+2 ≤ 0 [-2,-1] =(x-1)^2+2

Transcript text: $\frac{+1}{+2} \leq 0$ $[-2,-1]$ \[ =(x-1)^{2}+2 \]

Solution

Solution Steps

Solution Approach

To solve the inequality $\frac{x+1}{x+2} \leq 0$, we need to find the values of $x$ for which the fraction is less than or equal to zero. This involves finding the critical points where the numerator and denominator are zero and analyzing the sign changes around these points.
For the function $f(x) = (x-1)^2 + 2$, we need to determine the axis of symmetry, the domain, and the range. The axis of symmetry for a quadratic function $ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. The domain of a quadratic function is all real numbers, and the range can be found by identifying the vertex of the parabola.

Step 1: Solve the Inequality

To solve the inequality

\[ \frac{x + 1}{x + 2} \leq 0, \]

we find the critical points where the numerator and denominator are zero. The critical points are $x = -1$ (numerator) and $x = -2$ (denominator). The solution set for the inequality is

\[ (-2, -1). \]

Step 2: Analyze the Quadratic Function

For the quadratic function

\[ f(x) = (x - 1)^2 + 2, \]

we determine the axis of symmetry, which is given by

\[ x = 1. \]

Step 3: Determine the Domain and Range

The domain of the function is

\[ \mathbb{R} \quad \text{(all real numbers)}. \]

To find the range, we evaluate the minimum value of the function at the vertex $x = 1$:

\[ f(1) = (1 - 1)^2 + 2 = 2. \]

Thus, the range of the function is