\( g(x) \) is the following function in \( x \in \mathbb{R} \) : \[ g(x)=x^{2}+8 x-6 \] Find the value of \( g(-5) \). \[ (-5)^{2}+8(-5)-6=25-40-6=- \] Use calculus to find the value of \( x \) which gives the minimum value of \( g(x) \).

To determine the value of \( x \) that gives the minimum value of the function \( g(x) = x^2 + 8x - 6 \) using calculus, follow these steps: ### 1. **Find the Derivative of \( g(x) \):** The derivative of a function gives the rate at which the function is changing at any point \( x \). For \( g(x) \): \[ g(x) = x^2 + 8x - 6 \] Differentiate \( g(x) \) with respect to \( x \): \[ g'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(8x) - \frac{d}{dx}(6) = 2x + 8 \] ### 2. **Set the Derivative Equal to Zero to Find Critical Points:** To find the critical points (which can be minima, maxima, or saddle points), set the derivative equal to zero and solve for \( x \): \[ g'(x) = 2x + 8 = 0 \] \[ 2x = -8 \] \[ x = -4 \] ### 3. **Determine if the Critical Point is a Minimum:** Since the coefficient of \( x^2 \) in the original function \( g(x) \) is positive (\( 1 \)), the parabola opens upwards. This means that the critical point at \( x = -4 \) is indeed a **minimum**. ### 4. **Conclusion:** The value of \( x \) that gives the minimum value of \( g(x) \) is: \[ \boxed{-4} \] ### **Verification (Optional):** To confirm, you can plug \( x = -4 \) back into the original function: \[ g(-4) = (-4)^2 + 8(-4) - 6 = 16 - 32 - 6 = -22 \] Thus, the minimum value of \( g(x) \) is \(-22\) at \( x = -4 \).