\begin{tabular}{l}\( \qquad g(x)=x^{2}+8 x-6 \) \\ \( \qquad(-5)^{2}+8(-5)-6=25-40 \sim 6=-2 \) \\ Find the vallowing function in \( x \in \mathbb{R} \) : \\ \( \qquad g(-5) \) (c) \\ Use calculus to find the value of \( x \) which gives the minimum value of \( g(x) \) \\ \hline\end{tabular}

To find the value of \( x \) that gives the **minimum** of the function \( g(x) = x^{2} + 8x - 6 \) using calculus, follow these steps: ### 1. **Find the First Derivative** The first derivative of \( g(x) \) with respect to \( x \) will help us determine the critical points where the function could have minima or maxima. \[ g'(x) = \frac{d}{dx}(x^{2} + 8x - 6) = 2x + 8 \] ### 2. **Set the First Derivative to Zero** To find the critical points, set the first derivative equal to zero and solve for \( x \): \[ 2x + 8 = 0 \\ 2x = -8 \\ x = -4 \] ### 3. **Determine if It's a Minimum** To confirm that \( x = -4 \) is indeed a point of minimum, check the second derivative: \[ g''(x) = \frac{d}{dx}(2x + 8) = 2 \] Since \( g''(x) = 2 \) is **positive**, the function is concave upward at \( x = -4 \), indicating that this point is a **minimum**. ### 4. **Find the Minimum Value of \( g(x) \)** (Optional) If you need to find the minimum value of the function, substitute \( x = -4 \) back into the original function: \[ g(-4) = (-4)^{2} + 8(-4) - 6 = 16 - 32 - 6 = -22 \] ### **Conclusion** The function \( g(x) = x^{2} + 8x - 6 \) attains its **minimum** value at: \[ \boxed{x = -4} \] This means that \( x = -4 \) is the value in \( \mathbb{R} \) where \( g(x) \) reaches its lowest point.