A graphing device is recommended. Graph the polynomial in the given viewing rectangle. \[ y=x^{5}-6 x^{2}+4,[-3,3] \text { by }[-5,10] \] Find the coordinates of all local extrema, rounded to two decimal places. (If an answer does not exist, enter DwE) local minimum \( (x, y)=(\square) \) local maximum \( \quad(x, y)=( \) State the domain and range. (Enter your answers using interval notation. Round your answers to two decimal places.) domain range

To analyze the polynomial \( y = x^5 - 6x^2 + 4 \) and find its local extrema, we first need to compute its derivative and find the critical points. ### Step 1: Find the derivative The derivative of the polynomial is: \[ y' = 5x^4 - 12x \] ### Step 2: Set the derivative to zero To find the critical points, we set the derivative equal to zero: \[ 5x^4 - 12x = 0 \] Factoring out \( x \): \[ x(5x^3 - 12) = 0 \] This gives us one critical point at \( x = 0 \). Now we solve \( 5x^3 - 12 = 0 \): \[ 5x^3 = 12 \quad \Rightarrow \quad x^3 = \frac{12}{5} \quad \Rightarrow \quad x = \sqrt[3]{\frac{12}{5}} \approx 1.24 \] We also have the negative root: \[ x = -\sqrt[3]{\frac{12}{5}} \approx -1.24 \] ### Step 3: Find the second derivative To determine whether these critical points are local minima or maxima, we compute the second derivative: \[ y'' = 20x^3 - 12 \] ### Step 4: Evaluate the second derivative at the critical points 1. For \( x = 0 \): \[ y''(0) = 20(0)^3 - 12 = -12 \quad (\text{local maximum}) \] 2. For \( x \approx 1.24 \): \[ y''(1.24) = 20(1.24)^3 - 12 \approx 20(1.91) - 12 \approx 38.2 - 12 = 26.2 \quad (\text{local minimum}) \] 3. For \( x \approx -1.24 \): \[ y''(-1.24) = 20(-1.24)^3 - 12 \approx 20(-1.91) - 12 \approx -38.2 - 12 = -50.2 \quad (\text{local maximum}) \] ### Step 5: Find the coordinates of the local extrema Now we find the \( y \)-values at these critical points: 1. For \( x = 0 \): \[ y(0) = 0^5 - 6(0)^2 + 4 = 4 \quad \Rightarrow \quad (0, 4) \quad (\text{local maximum}) \] 2. For \( x \approx 1.24 \): \[ y(1.24) = (1.24)^5 - 6(1.24)^2 + 4 \approx 2.43 - 9.14 + 4 \approx -2.71 \quad \Rightarrow \quad (1.24, -2.71) \quad (\text{local minimum}) \] 3. For \( x \approx -1.24 \): \[ y(-1.24) = (-1.24)^5 - 6(-1.24)^2 + 4 \approx -2.43 - 9.14 + 4 \approx -7.57 \quad \Rightarrow \quad (-1.24, -7.57) \quad (\text{local maximum}) \] ### Summary of Local Extrema - Local minimum: \( (1.24, -2.71) \) - Local maximum: \( (0, 4) \) ### Step 6: Domain and Range - The domain of the polynomial is all real numbers: \[ \text{Domain: } (-\infty, \infty) \] - The range can be determined from the graph. The local minimum is approximately -2.71 and the local maximum is 4. The polynomial approaches \( \infty \) as \( x \to \infty \) and \( -\infty \) as \( x \to -\infty \). Thus, the range is: \[ \text{Range: } (-\infty, 4] \cup [-2.71, \infty) \] ### Final Answers - Local minimum: \( (x, y) = (1.24, -2.71) \) - Local maximum: \( (x, y) = (0, 4