The problem in the image is asking for the value(s) of $c$ that satisfy the Mean Value Theorem (MVT) for the given piecewise function $g(x)$ on the interval $[-7, 14]$.

Steps to solve:

The Mean Value Theorem (MVT) states that, for a function continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, there exists at least one $c$ in $(a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

1. Identifying the function $g(x)$

• For $-7 \leq x \leq 0$, $g(x) = x^3$
• For $0 < x \leq 14$, $g(x) = x^2$

We will apply the Mean Value Theorem to both intervals separately. Let's break it down:

Interval 1: $-7 \leq x \leq 0$

Here, $g(x) = x^3$.

• $g(-7) = (-7)^3 = -343$
• $g(0) = (0)^3 = 0$

Using the MVT:

$g'(c_1) = \frac{g(0) - g(-7)}{0 - (-7)} = \frac{0 - (-343)}{7} = \frac{343}{7} = 49$

Since $g(x) = x^3$, we differentiate it:

$g'(x) = 3x^2$

Now, we solve for $c_1$:

$3c_1^2 = 49 \quad \Rightarrow \quad c_1^2 = \frac{49}{3} \quad \Rightarrow \quad c_1 = \pm \frac{\sqrt{49}}{\sqrt{3}} = \pm \frac{7}{\sqrt{3}} = \pm \frac{7\sqrt{3}}{3}$

Since $c_1$ must lie within the interval $[-7, 0]$, we select the negative value:

$c_1 = -\frac{7\sqrt{3}}{3}$

Interval 2: $0 \leq x \leq 14$

Here, $g(x) = x^2$.

• $g(0) = 0^2 = 0$
• $g(14) = (14)^2 = 196$

Using the MVT:

$g'(c_2) = \frac{g(14) - g(0)}{14 - 0} = \frac{196 - 0}{14} = 14$

Since $g(x) = x^2$, we differentiate it:

$g'(x) = 2x$

Now, we solve for $c_2$:

$2c_2 = 14 \quad \Rightarrow \quad c_2 = 7$

Final answer:

The values of $c$ are:

$c = -\frac{7\sqrt{3}}{3}, 7$

Would you like a further breakdown of any of these steps?

Here are 5 related questions to explore:

1. What conditions must a function meet for the Mean Value Theorem to apply?
2. How does the Mean Value Theorem differ from Rolle's Theorem?
3. Can you apply the Mean Value Theorem if the function is not continuous?
4. How does the choice of interval affect the value of $c$?
5. How do piecewise functions affect the application of the Mean Value Theorem?

Tip: When solving problems involving the MVT, always verify both continuity and differentiability over the specified intervals.