Let $$ f(x) $$ be a function that is differentiable on the entire real line. If $$ f(-6)=-10 $$ and $$ f(3)=8 $$, we can guarantee that there is a point $$ c $$ in the interval $$ (-6,3) $$ such that $$ f^{\prime}(c)= $$ We know that such a point exists by the Squeeze Theorem Intermediate Value Theorem Mean value Theorem

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Given that $$ f(x) $$ is differentiable on the entire real line, and you have the points $$ f(-6) = -10 $$ and $$ f(3) = 8 $$, you can apply the **Mean Value Theorem (MVT)** to guarantee the existence of a point $$ c $$ in the interval $$(-6, 3)$$ where the derivative $$ f'(c) $$ equals the average rate of change over the interval.

**Mean Value Theorem Statement:**
If a function $$ f $$ is continuous on the closed interval $$[a, b]$$ and differentiable on the open interval $$(a, b)$$, then there exists at least one point $$ c $$ in $$(a, b)$$ such that:
$$
f'(c) = \frac{f(b) - f(a)}{b - a}
$$

**Applying this to your problem:**
- $$ a = -6 $$
- $$ b = 3 $$
- $$ f(a) = f(-6) = -10 $$
- $$ f(b) = f(3) = 8 $$

Compute the average rate of change:
$$
f'(c) = \frac{f(3) - f(-6)}{3 - (-6)} = \frac{8 - (-10)}{9} = \frac{18}{9} = 2
$$

Therefore, there exists a point $$ c $$ in $$(-6, 3)$$ such that:
$$
f'(c) = 2
$$

**Conclusion:**
This result is guaranteed by the **Mean Value Theorem**.

**Answer:**  
Such a point exists by the Mean Value Theorem.
There exists a point $$ c $$ in $$(-6, 3)$$ where $$ f'(c) = 2 $$, guaranteed by the Mean Value Theorem.

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