Let's work through the properties **step by step**. We'll analyze and build the function \( f(x) \) that fits all the given requirements. --- ## **Step 1: Domain** **Domain:** \( (-\infty, -2) \cup (-2, \infty) \) - The function is undefined at \( x = -2 \). - The function is defined everywhere else. --- ## **Step 2: Range** **Range:** \( (-5, \infty) \) - The function can take on any value greater than \(-5\) but **not** \(-5\). - It never reaches \(-5\), but it can get arbitrarily close. --- ## **Step 3: Vertical Asymptote** **Vertical Asymptote at \( x = -2 \):** - As \( x \to -2^+ \) or \( x \to -2^- \), \( f(x) \to \pm \infty \). --- ## **Step 4: End Behavior / Horizontal Asymptotes** **Limits at infinity:** - \( \lim_{x \to -\infty} f(x) = 2 \) - \( \lim_{x \to \infty} f(x) = -5 \) So: - As \( x \to -\infty \), function approaches \( 2 \). - As \( x \to \infty \), function approaches \( -5 \) (but never equals). --- ## **Step 5: Limit and Discontinuity at \( x = 3 \)** - \( \lim_{x \to 3} f(x) = 2 \) - \( f(x) \) is **discontinuous** at \( x = 3 \) This means either a jump, removable, or infinite discontinuity at \( x=3 \). --- ## **Step 6: Increasing/Decreasing Intervals** - \( f'(x) > 0 \) on \( (-\infty, -2) \): Increasing to the left of \( x = -2 \) - \( f'(x) < 0 \) on \( (-2, 3) \): Decreasing from \( x = -2 \) to \( x = 3 \) --- ## **Step 7: \( f'(4) \) Not Defined, but \( f(x) \) Continuous at 4** - Sharp corner or cusp at \( x=4 \), but no break in the graph. --- ### **Let's Build the Function** Let’s try to construct a function that fits these properties. We'll start with a basic rational transformation and then tweak as needed. #### **Base function idea:** A function with a vertical asymptote at \( x = -2 \) and horizontal asymptotes at \( y = 2 \) as \( x \to -\infty \), and \( y = -5 \) as \( x \to \infty \), and range \( (-5, \infty) \). A rational function like: \[ f(x) = \frac{a}{x+2} + b \] But this function only has one horizontal asymptote. To get **two different asymptotes** at the ends, we need a rational function where numerator and denominator have the same degree. Let’s set: \[ f(x) = \frac{m x + n}{x + 2} \] Let's find \( m, n \) so that: - As \( x \to \infty \): \( f(x) \to m \) (since the degree is the same, divide coefficients). - As \( x \to -\infty \): \( f(x) \to m \) (same as above). But we want **different limits at infinity**. We need to tweak further. Try: \[ f(x) = \frac{A x + B}{x+2} + C \] But this still only allows one horizontal asymptote. #### **Key Rational Function for Two Horizontal Asymptotes** Let’s try this standard form: \[ f(x) = \frac{ax + b}{cx + d} \] As \( x \to \infty \), \( f(x) \to \frac{a}{c} \). As \( x \to -\infty \), \( f(x) \to \frac{a}{c} \) (since the degrees are the same). But we want **different horizontal asymptotes**. This can be achieved by using a function like: \[ f(x) = \frac{P(x)}{Q(x)} + R \] Or by shifting and stretching. Alternatively, a function such as \[ f(x) = \frac{N(x)}{D(x)} \] where the numerator and denominator are both degree 2. Suppose: \[ f(x) = \frac{a x^2 + b x + c}{x+2} \] But then as \( x \to \infty \), numerator dominates, so \( f(x) \to \infty \). So instead, try: \[ f(x) = \frac{a x + b}{x + 2} + d \] But as above, this only gives a single horizontal asymptote. #### **Let's Try an Exponential Decay** Alternatively, use a function with a vertical asymptote at \( x = -2 \) and horizontal asymptotes at different ends. A function like: \[ f(x) = -5 + \frac{A}{x+2} \] But as \( x \to \infty \), \( f(x) \to -5 \), as \( x \to -\infty \), \( f(x) \to -5 \). Not enough. #### **Solution: Use a Shifted Reciprocal Function** Let’s look for a function that increases toward \( 2 \) as \( x \to -\infty \), and decreases toward \( -5 \) as \( x \to \infty \), and has vertical asymptote at \( x=-2 \). Let’s set: \[ f(x) = \frac{K}{x+2} + L \] Let’s find \( K \) and \( L \). As \( x \to -\infty \), \( f(x) \approx L \) As \( x \to \infty \), \( f(x) \approx L \) Still both ends go to \( L \). But if we introduce a negative sign and a shift: \[ f(x) = a + \frac{b}{x+2} \] Same problem. Let’s try a function like: \[ f(x) = -5 + \frac{7}{1 + e^{x+2}} \] But as \( x \to \infty \), \( e^{x+2} \to \infty \Rightarrow f(x) \to -5 \). As \( x \to -\infty \), \( e^{x+2} \to 0 \Rightarrow f(x) \to -5 + 7 = 2 \). **This matches the behavior!** - Vertical asymptote at \( x = -2 \)? Let's check the denominator at \( x=-2 \): \[ 1 + e^{-2 + 2} = 1 + e^{0} = 2 \neq 0 \] **So this function is continuous at \( x = -2 \), not a vertical asymptote.** #### **Try a Negative Exponent:** \[ f(x) = -5 + 7 \cdot e^{-(x+2)} \] As \( x \to \infty \), \( e^{-(x+2)} \to 0 \Rightarrow f(x) \to -5 \) As \( x \to -\infty \), \( e^{-(x+2)} \to \infty \Rightarrow f(x) \to \infty \) So, not quite. #### **Try Rational Function with Quadratic Denominator** Let’s try: \[ f(x) = \frac{A}{x + 2} + B \] We already tried this. Try: \[ f(x) = \frac{C}{x + 2} + D \] Let’s force the limits: - As \( x \to -\infty \), \( f(x) \to D \) - As \( x \to \infty \), \( f(x) \to D \) Not sufficient. #### **A Better Approach: Use Reciprocal with Different Numerator** Try: \[ f(x) = \frac{ax + b}{x+2} \] Find \( a, b \) so that: - As \( x \to \infty \), \( f(x) \to a \) (since degrees are the same) - As \( x \to -\infty \), \( f(x) \to a \) Still both ends same. #### **Try a Rational Function with Horizontal Asymptotes at Different Ends** A function like \[ f(x) = \frac{p}{1 + e^{x}} \] has horizontal asymptotes at \( y = p \) and \( y = 0 \). So, we can **combine two horizontal