The table gives selected values of the differentiable function \[h\].

\[x\]\[-5\]\[-4\]\[-3\]\[-2\]

\[h(x)\]

\[18\]

\[13\]

\[15\]

\[19\]

Below is Tom's attempt to write a formal justification for the fact that there exists a value \[c\] in the interval \[(-4,-3)\] such that \[h'(c)=2\].

**Is Tom's justification complete? If not, why?**

Tom's justification:

We are given that \[h\] is differentiable, which means it's both differentiable and continuous over the interval \[[-4,-3]\]. Furthermore, \[\dfrac{h(-3)-h(-4)}{-3-(-4)}=2\].

So, according to the mean value theorem, there exists a value \[c\] somewhere in the interval \[(-4,-3)\] such that \[h'(c)=2\].

Tom's justification relies on the **Mean Value Theorem (MVT)** to show that there exists a point \( c \) in the interval \( (-4, -3) \) where the derivative \( h'(c) = 2 \). To evaluate if Tom's justification is complete, let's review the MVT requirements and Tom's application of it.

### Mean Value Theorem (MVT) Conditions:
1. **Continuity**: The function \( h(x) \) must be continuous on the closed interval \([a, b]\).
2. **Differentiability**: The function \( h(x) \) must be differentiable on the open interval \((a, b)\).

If these conditions are met, then there exists at least one point \( c \) in \((a, b)\) such that:
\[
h'(c) = \frac{h(b) - h(a)}{b - a}
\]

### Applying MVT to Tom's Situation:

- **Function Differentiability**: Tom correctly states that since \( h \) is differentiable, it implies \( h \) is also continuous on \([-4, -3]\), satisfying the conditions of the MVT.

- **Calculating the Slope**:
    - Tom calculates the average rate of change of \( h(x) \) over the interval \([-4, -3]\) as follows:
      \[
      \frac{h(-3) - h(-4)}{-3 - (-4)} = \frac{15 - 13}{-3 + 4} = \frac{2}{1} = 2
      \]
    - This calculation is correct, and the result is \( 2 \).

- **Conclusion Using MVT**: Given that \( h \) is continuous and differentiable on the specified intervals, and the average rate of change is \( 2 \), MVT guarantees that there is at least one point \( c \) in \((-4, -3)\) such that \( h'(c) = 2 \).

### Is Tom's Justification Complete?

**Yes, Tom's justification is complete and correct.** 

He correctly:
1. Verified that \( h \) meets the requirements for the Mean Value Theorem (continuity and differentiability).
2. Correctly applied the MVT by finding the average rate of change and equating it to the derivative at some point \( c \) in the interval.

Tom's reasoning demonstrates a sound understanding of how to apply the Mean Value Theorem in this context.

### Follow-up Questions:

1. What does the Mean Value Theorem tell us about the relationship between differentiability and continuity?
2. How would you justify the existence of \( c \) if the function were only continuous but not differentiable?
3. Can the Mean Value Theorem be applied to any differentiable function over any interval?
4. What is the geometric interpretation of the Mean Value Theorem?
5. How would the argument change if \( h'(c) = 0 \) instead of \( 2 \)?

**Tip:** Remember, the Mean Value Theorem provides a link between the average rate of change over an interval and the instantaneous rate of change at a point within the interval, highlighting the importance of differentiability and continuity.

Explore how the Mean Value Theorem guarantees the existence of a point where the derivative of a differentiable function equals the average rate of change over an interval. Includes a detailed analysis of Tom's justification for a specific function h(x) over the interval (-4, -3).

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