To solve this problem, we need to follow these steps:

Step 1: Calculate the Mean Slope of $f(x)$ on the Interval $[-5, 3]$

The mean (or average) slope of $f(x)$ on the interval $[-5, 3]$ is given by: $\frac{f(3) - f(-5)}{3 - (-5)}$ The function is: $f(x) = 1 - 2x^2$

Calculate $f(3)$ and $f(-5)$:

1. $f(3) = 1 - 2(3)^2 = 1 - 18 = -17$
2. $f(-5) = 1 - 2(-5)^2 = 1 - 2 \times 25 = 1 - 50 = -49$

Substitute these values into the mean slope formula:

$\frac{f(3) - f(-5)}{3 - (-5)} = \frac{-17 - (-49)}{3 + 5} = \frac{-17 + 49}{8} = \frac{32}{8} = 4$

So, the mean slope is $4$.

Step 2: Apply the Mean Value Theorem (MVT)

According to the Mean Value Theorem, there exists a point $c$ in the interval $(-5, 3)$ such that $f'(c)$ is equal to this mean slope.

1. Find $f'(x)$: Differentiate $f(x) = 1 - 2x^2$: $f'(x) = -4x$

2. Set $f'(c) = 4$ and solve for $c$: $-4c = 4$ $c = -1$

Conclusion

The point $c$ that satisfies the Mean Value Theorem is $c = -1$.

Final Answer

• Mean slope of $f(x)$ on $[-5, 3]$: $4$
• Value of $c$ where $f'(c) = 4$: $c = -1$

Would you like further details on any part of this solution?

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Here are some related questions:

1. What would happen if the interval were changed, say to $[-4, 2]$?
2. How does the Mean Value Theorem apply to other types of functions, like trigonometric ones?
3. Can the Mean Value Theorem have more than one solution $c$ in an interval?
4. How do we interpret the Mean Value Theorem geometrically on a graph?
5. What are other key theorems related to derivatives and slopes?

Tip: In calculus, the Mean Value Theorem not only helps find average slopes but also guarantees at least one point where the instantaneous rate of change equals the average rate over an interval.