The number of continous and derivable function(s) f(x) such that f(1) = -1 , f(4) = 7 and `f'(x) gt 3` for all x `in` R is are :

Options

A. 0
B. 1
C. 3
D. infinite

Explanation

To solve the problem, we need to find the number of continuous and differentiable functions \( f(x) \) that satisfy the following conditions:

1. \( f(1) = -1 \)
2. \( f(4) = 7 \)
3. \( f'(x) > 3 \) for all \( x \in \mathbb{R} \)

### Step-by-Step Solution:

**Step 1: Apply the Mean Value Theorem**

According to the Mean Value Theorem (MVT), if a function \( f(x) \) 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:

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

In our case, let \( a = 1 \) and \( b = 4 \).

**Step 2: Calculate \( f'(c) \)**

Substituting the values into the MVT formula:

\[ f'(c) = \frac{f(4) - f(1)}{4 - 1} = \frac{7 - (-1)}{4 - 1} = \frac{8}{3} \]

**Step 3: Analyze the Result**

We find that \( f'(c) = \frac{8}{3} \). Now, we need to compare this result with the condition given in the problem, which states that \( f'(x) > 3 \) for all \( x \in \mathbb{R} \).

**Step 4: Compare \( \frac{8}{3} \) with 3**

Calculating \( \frac{8}{3} \):

\[ \frac{8}{3} \approx 2.67 \]

Since \( 2.67 < 3 \), this means that there is a contradiction. The derivative at some point \( c \) (which is \( \frac{8}{3} \)) does not satisfy the condition \( f'(x) > 3 \).

**Step 5: Conclusion**

Since we have found that the condition \( f'(x) > 3 \) cannot be satisfied given the values of \( f(1) \) and \( f(4) \), we conclude that there are no functions \( f(x) \) that meet all the given criteria.

Thus, the number of continuous and differentiable functions \( f(x) \) that satisfy the conditions is:

\[ \boxed{0} \]