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?

Answer 1

The answer is #c=1.75#

The mean value theorem states that if a function #f(x)# is continous on the interval #[a,b]# and differentiable on the interval #(a,b)#, then there exists a point #c# in the interval such that
#f'(c)=(f(b)-f(a))/(b-a)#

Here,

#f(x)=-(-2x+6)^(1/2)=-sqrt(-2x+6)#
The domain is #x in (-oo, 3]#
The function is continuous and differentiable on the interval #(-2,3)#

Therefore,

#f'(x)=-1/(2sqrt(-2x+6))*-2=1/sqrt(-2x+6)#
#f'(c)=1/sqrt(-2c+6)#
#f(-2)=-sqrt(4+6)=-sqrt10#
#f(3)=-sqrt(-6+6)=0#

Therefore,

#1/sqrt(-2c+6)=(0-(-sqrt10))/(3-(-2))#
#=>#, #1/sqrt(-2c+6)=sqrt10/5#
#=>#, #sqrt(-2c+6)=5/sqrt10#

Squaring both sides

#=>#, #-2c+6=25/10=2.5#
#2c=6-2.5=3.5#
#c=3.5/2=1.75#
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Answer 2

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.

  1. 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]).

  1. 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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Answer from HIX Tutor

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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