The function f, of, x, equals, x, cubed, minus, 10, x, squared, plus, 17, x, plus, 48*f*(*x*)=*x*3−10*x*2+17*x*+48 is graphed below. Plot a line segment connecting the points on f*f* where x, equals, minus, 1*x*=−1 and x, equals, 7, .*x*=7. Afterwards, determine all values of c*c* which satisfy the conclusion of the Mean Value Theorem for f*f* on the closed interval minus, 1, is less than or equal to, x, is less than or equal to, 7−1≤*x*≤7.

To solve this problem, we will first evaluate the function \( f(x) = x^3 - 10x^2 + 17x + 48 \) at the endpoints of the interval \([-1, 7]\), then apply the Mean Value Theorem (MVT).

### Step 1: Evaluate the function at the endpoints

1. **Calculate \( f(-1) \)**:
   \[
   f(-1) = (-1)^3 - 10(-1)^2 + 17(-1) + 48
   \]
   \[
   = -1 - 10 - 17 + 48 = 20
   \]

2. **Calculate \( f(7) \)**:
   \[
   f(7) = 7^3 - 10(7^2) + 17(7) + 48
   \]
   \[
   = 343 - 490 + 119 + 48 = 20
   \]

### Step 2: Mean Value Theorem

The Mean Value Theorem states that there exists at least one \( c \) in the interval \((a, b)\) such that:
\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]
Here, \( a = -1 \) and \( b = 7 \).

3. **Calculate the average rate of change**:
   \[
   f'(c) = \frac{f(7) - f(-1)}{7 - (-1)} = \frac{20 - 20}{8} = 0
   \]

### Step 3: Find \( c \) where \( f'(c) = 0 \)

4. **Calculate the derivative \( f'(x) \)**:
   \[
   f'(x) = 3x^2 - 20x + 17
   \]

5. **Set the derivative equal to zero**:
   \[
   3x^2 - 20x + 17 = 0
   \]

### Step 4: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = -20 \), \( c = 17 \).

\[
b^2 - 4ac = (-20)^2 - 4(3)(17) = 400 - 204 = 196
\]
\[
\sqrt{196} = 14
\]
\[
x = \frac{20 \pm 14}{6}
\]

Calculating the two solutions:
1. \( x_1 = \frac{34}{6} = \frac{17}{3} \approx 5.67 \)
2. \( x_2 = \frac{6}{6} = 1 \)

### Step 5: Conclusion

Thus, the values of \( c \) in the interval \([-1, 7]\) that satisfy the conclusion of the Mean Value Theorem are:
\[
c = 1 \quad \text{and} \quad c = \frac{17}{3}
\]

### Summary

- Points for the line segment: \( (-1, 20) \) and \( (7, 20) \).
- Values of \( c \) satisfying MVT: \( 1 \) and \( \frac{17}{3} \).

Would you like more details on any specific part of this solution, or do you have any questions? Here are some related questions to consider:

1. What is the significance of the Mean Value Theorem in calculus?
2. Can you explain how to derive a cubic function from its graph?
3. How would you find critical points of a function?
4. What is the role of the first derivative test in identifying local extrema?
5. Can you graph the function \( f(x) \) along with the line segment connecting the specified points? 

**Tip:** Always check if the endpoints yield the same value to verify applicability of the MVT!

The function f(x) = x^3 - 10x^2 + 17x + 48 is graphed below. Plot a line segment connecting the points on f where x = -1 and x = 7. Afterwards, determine all values of c which satisfy the conclusion of the Mean Value Theorem for f on the closed interval -1 ≤ x ≤ 7.

Explore the application of the Mean Value Theorem for the cubic function f(x) = x^3 - 10x^2 + 17x + 48. This solution includes evaluating the function at critical points and determining values of c in the interval -1 ≤ x ≤ 7 that satisfy the theorem. Suitable for advanced high school students.

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