How do you determine whether #f(x) = sqrt( 2x + 3 ) 1# satisfies the hypotheses of the Mean Value Theorem on the interval [3,11] and find all value(s) of c that satisfy the conclusion of the theorem?
The Mean Value Theorem has two hypotheses:
To do the additional algebra problem,
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To determine whether ( f(x) = \sqrt{2x + 3}  1 ) satisfies the hypotheses of the Mean Value Theorem on the interval ([3,11]), you need to verify that the function is continuous on the closed interval ([3,11]) and differentiable on the open interval ((3,11)).

Check continuity:
 ( f(x) ) is continuous on the closed interval ([3,11]) because it is a composition of continuous functions.

Check differentiability:
 ( f(x) ) is differentiable on the open interval ((3,11)) because it is a composition of differentiable functions.
Since ( f(x) ) satisfies the hypotheses of the Mean Value Theorem on the interval ([3,11]), you can apply the theorem to find the value(s) of ( c ) that satisfy the conclusion.
The Mean Value Theorem states that there exists a ( c ) in the open interval ((3,11)) such that:
[ f'(c) = \frac{f(11)  f(3)}{11  3} ]
Now, find ( f'(x) ) and evaluate ( f(11) ) and ( f(3) ):
[ f'(x) = \frac{d}{dx} \left(\sqrt{2x + 3}  1\right) = \frac{1}{2\sqrt{2x + 3}} ] [ f(11) = \sqrt{2(11) + 3}  1 = \sqrt{25}  1 = 4  1 = 3 ] [ f(3) = \sqrt{2(3) + 3}  1 = \sqrt{9}  1 = 3  1 = 2 ]
Plug these values into the Mean Value Theorem equation:
[ f'(c) = \frac{3  2}{11  3} = \frac{1}{8} ]
Now, solve for ( c ):
[ \frac{1}{2\sqrt{2c + 3}} = \frac{1}{8} ]
[ 2\sqrt{2c + 3} = 8 ]
[ \sqrt{2c + 3} = 4 ]
[ 2c + 3 = 16 ]
[ 2c = 13 ]
[ c = \frac{13}{2} ]
So, ( c = \frac{13}{2} ) satisfies the conclusion of the Mean Value Theorem.
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