Given the function #f(x)=-(-2x+6)^(1/2)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-2,3] and find the c?
The answer is
Here,
Therefore,
Therefore,
Squaring both sides
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To determine if ( f(x) = -\sqrt{-2x + 6} ) satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval ([-2, 3]), first, check if the function is continuous on the interval ([-2, 3]). Then, check if the function is differentiable on the open interval ((-2, 3)). If both conditions are satisfied, then the Mean Value Theorem applies.
- Continuity: Check if the function is continuous on the closed interval ([-2, 3]). Since square root functions are continuous everywhere their radicands are defined, ( -2x + 6 ) must be non-negative on ([-2, 3]).
Solve for when ( -2x + 6 = 0 ): [ -2x + 6 = 0 ] [ 2x = 6 ] [ x = 3 ]
Since ( -2x + 6 ) is a linear function with a negative slope, it goes from positive to negative at ( x = 3 ), so it's non-negative on the interval ([-2, 3]).
- Differentiability: Check if the function is differentiable on the open interval ((-2, 3)). Since the function is a square root function, it's differentiable wherever its radicand is differentiable. The radicand ( -2x + 6 ) is a linear function and is differentiable everywhere.
Therefore, the hypotheses of the Mean Value Theorem are satisfied on the interval ([-2, 3]).
To find ( c ), which is the point guaranteed by the Mean Value Theorem, first, find the average rate of change of ( f ) on the interval ([-2, 3]): [ \frac{f(3) - f(-2)}{3 - (-2)} ]
[ \frac{-\sqrt{-2(3) + 6} - (-\sqrt{-2(-2) + 6})}{3 - (-2)} ] [ \frac{-\sqrt{0} - (-\sqrt{10})}{5} ] [ \frac{0 - (-\sqrt{10})}{5} ] [ \frac{\sqrt{10}}{5} ]
Next, find the derivative of ( f(x) ): [ f'(x) = \frac{-1}{2\sqrt{-2x + 6}} ]
Then, find the value of ( c ) where ( f'(c) ) equals the average rate of change: [ f'(c) = \frac{\sqrt{10}}{5} ]
[ \frac{-1}{2\sqrt{-2c + 6}} = \frac{\sqrt{10}}{5} ]
Solve for ( c ): [ \frac{-1}{2\sqrt{-2c + 6}} = \frac{\sqrt{10}}{5} ] [ -1 = 2\sqrt{-2c + 6}\cdot \frac{\sqrt{10}}{5} ] [ -\frac{5}{2\sqrt{10}} = \sqrt{-2c + 6} ] [ \frac{25}{40} = -2c + 6 ] [ \frac{5}{8} = -2c + 6 ] [ -\frac{35}{8} = -2c ] [ c = \frac{35}{16} ]
Therefore, ( c = \frac{35}{16} ).
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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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