Given the function # f(x)=1/sqrt(x+2)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-1,23] and find the c?
The conclusion of the MVT is
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To determine if ( f(x) = \frac{1}{\sqrt{x + 2}} ) satisfies the hypotheses of the Mean Value Theorem on the interval ([-1, 23]) and find the ( c ), we need to check two conditions:
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Continuity: Check if ( f(x) ) is continuous on ([-1, 23]). Since ( f(x) ) is a rational function and is continuous on its domain, which is ( (-2, \infty) ), it is also continuous on ([-1, 23]) since ([-1, 23]) is a subset of its domain.
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Differentiability: Check if ( f(x) ) is differentiable on ((-1, 23)). The derivative of ( f(x) ) is given by:
[ f'(x) = \frac{-1}{2(x + 2)^{3/2}} ]
This derivative exists for all ( x ) in ((-2, \infty)), which includes ([-1, 23]), so ( f(x) ) is differentiable on ([-1, 23]).
Since ( f(x) ) is continuous and differentiable on ([-1, 23]), it satisfies the hypotheses of the Mean Value Theorem.
To find the value of ( c ), we can use the Mean Value Theorem formula:
[ f'(c) = \frac{f(b) - f(a)}{b - a} ]
where ( a = -1 ), ( b = 23 ), and ( c ) is the value we need to find.
First, calculate ( f(-1) ) and ( f(23) ):
[ f(-1) = \frac{1}{\sqrt{-1 + 2}} = \frac{1}{\sqrt{1}} = 1 ] [ f(23) = \frac{1}{\sqrt{23 + 2}} = \frac{1}{\sqrt{25}} = \frac{1}{5} ]
Now, substitute these values into the Mean Value Theorem formula:
[ f'(c) = \frac{\frac{1}{5} - 1}{23 - (-1)} = \frac{-4/5}{24} = -\frac{1}{30} ]
Finally, solve for ( c ):
[ -\frac{1}{30} = \frac{-1}{2(c + 2)^{3/2}} ] [ (c + 2)^{3/2} = 30 ] [ c + 2 = 30^{2/3} ] [ c + 2 = 10 ] [ c = 8 ]
So, the value of ( c ) that satisfies the Mean Value Theorem on ([-1, 23]) is ( c = 8 ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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