If f(0) = 4, and f'(x) = 3/(x^2 + 2), the lower bound of f(2) estimated by mean value theorem is

Steps to Solve

1. Identify Given Information

The provided information includes:

• ( f(0) = 4 )
• The derivative ( f'(x) = \frac{3}{x^2 + 2} )

2. Apply the Mean Value Theorem (MVT)

According to the Mean Value Theorem, if ( f ) is continuous on ([a,b]) and differentiable on ((a,b)), then there exists a ( c ) in ((a,b)) such that: $$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

Here, let ( a = 0 ) and ( b = 2 ). Thus, $$ f'(c) = \frac{f(2) - f(0)}{2 - 0} $$

3. Calculate the Derivative at the Specific Points

The derivative ( f'(c) = \frac{3}{c^2 + 2} ). Setting the two expressions equal gives: $$ \frac{3}{c^2 + 2} = \frac{f(2) - 4}{2} $$

4. Find the Value of ( c )

Since ( c ) must be in the interval ((0, 2)), we can maximize ( f'(x) ) to find the minimum value for ( f'(c) ). This occurs at ( c = 2 ): $$ f'(2) = \frac{3}{2^2 + 2} = \frac{3}{4 + 2} = \frac{3}{6} = \frac{1}{2} $$

5. Substituting to find ( f(2) )

Now, substituting ( f'(c) ): $$ \frac{1}{2} = \frac{f(2) - 4}{2} $$ Multiplying both sides by 2 gives: $$ 1 = f(2) - 4 $$ Adding 4 to both sides results in: $$ f(2) = 5 $$

6. Establishing the Lower Bound

Thus, according to the Mean Value Theorem, the lower bound estimate for ( f(2) ) is 5.