Find the absolute extrema of the function on the closed interval. (If an answer does not exist, enter DN \[ g(x)=8\left(1+\frac{1}{x}+\frac{1}{x^{2}}\right), \quad[-5,4] \]

To find the absolute extrema of the function \( g(x) = 8\left(1 + \frac{1}{x} + \frac{1}{x^2}\right) \) on the closed interval \([-5, 4]\), we first need to examine the function for critical points and check the endpoints of the interval. 1. **Finding the derivative**: We first differentiate \( g(x) \): \[ g'(x) = 8\left( -\frac{1}{x^2} - \frac{2}{x^3} \right) = -8\left( \frac{1}{x^2} + \frac{2}{x^3} \right) = -\frac{8(1 + \frac{2}{x})}{x^3} \] Setting \( g'(x) = 0 \) means finding when the numerator equals zero, which does not provide a solution since \( 1 + \frac{2}{x} \neq 0 \) for \( x \neq -2 \). However, we see \( g'(x) \) is not defined at \( x = 0 \), which is outside the preferred interval \([-5, 4]\). 2. **Evaluating endpoints**: Now we check the values at the endpoints of the interval: \[ g(-5) = 8\left(1 + \frac{1}{-5} + \frac{1}{(-5)^2}\right) = 8\left(1 - \frac{1}{5} + \frac{1}{25}\right) = 8\left(\frac{25 - 5 + 1}{25}\right) = 8\left(\frac{21}{25}\right) = \frac{168}{25} = 6.72 \] \[ g(4) = 8\left(1 + \frac{1}{4} + \frac{1}{16}\right) = 8\left(1 + 0.25 + 0.0625\right) = 8\left(1.3125\right) = 10.5 \] 3. **Conclusion**: Since the function is always decreasing on the interval we analyzed (except at \( x = 0 \), which is outside the interval), we determine that: - The minimum occurs at \( x = -5 \): \( g(-5) = 6.72 \) - The maximum occurs at \( x = 4 \): \( g(4) = 10.5 \) Thus, the absolute minimum is \( 6.72 \) and the absolute maximum is \( 10.5 \). In conclusion, the absolute extrema of \( g(x) \) on the interval \([-5, 4]\) are: - Absolute Minimum: \( 6.72 \) at \( x = -5 \). - Absolute Maximum: \( 10.5 \) at \( x = 4 \).