\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}

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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.
The function $$ g(x) = x^{2} + 8x - 6 $$ reaches its minimum at $$ x = -4 $$.

Find the vallowing function in :

(c)

Use calculus to find the value of which gives the minimum value of

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