Find the value or values of c that satisfy the equation f(b) - f(a) / b - a = f'(c) in the conclusion of the mean value theorem for the given function and interval.

Steps to Solve

1. Identify the function and interval

The function given is ( f(x) = \sqrt{x - 1} ) and the interval is ([1, 8]).

2. Calculate ( f(a) ) and ( f(b) )

To apply the Mean Value Theorem (MVT):

• Let ( a = 1 )

$$ f(a) = f(1) = \sqrt{1 - 1} = \sqrt{0} = 0 $$

• Let ( b = 8 )

$$ f(b) = f(8) = \sqrt{8 - 1} = \sqrt{7} $$

3. Compute the difference quotient

Using the MVT formula: $$ \frac{f(b) - f(a)}{b - a} $$

Substituting in the values we found:

$$ \frac{f(8) - f(1)}{8 - 1} = \frac{\sqrt{7} - 0}{8 - 1} = \frac{\sqrt{7}}{7} $$

4. Find the derivative of ( f(x) )

To find ( c ), compute the derivative ( f'(x) ):

Using the power rule: $$ f(x) = (x - 1)^{1/2} $$

Then: $$ f'(x) = \frac{1}{2}(x - 1)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x - 1}} $$

5. Set the derivative equal to the difference quotient

Now set ( f'(c) = \frac{\sqrt{7}}{7} ):

$$ \frac{1}{2\sqrt{c - 1}} = \frac{\sqrt{7}}{7} $$

6. Solve for ( c )

Cross-multiply to solve for ( c ):

$$ 7 = 2\sqrt{7}\sqrt{c - 1} $$

Squaring both sides:

$$ 49 = 4 \cdot 7(c - 1) $$ $$ 49 = 28(c - 1) $$

Now, divide both sides by 28:

$$ c - 1 = \frac{49}{28} = \frac{7}{4} $$

Thus:

$$ c = \frac{7}{4} + 1 = \frac{7}{4} + \frac{4}{4} = \frac{11}{4} $$