How do you determine whether the function #f(x) = abs(x-3)# satisfies the hypotheses of the mean value theorem on the indicated interval (a,b), and if so how do you find all numbers c on [0,4] ?

Answer 1

The Mean Value Theorem has two hypotheses:

H1 : #f# is continuous on the closed interval #[a,b]#
H2 : #f# is differentiable on the open interval #(a,b)#.
In this question, #f(x) = abs(x-3)# , #a=0# and #b=4#.
We can apply the Mean Value Theorem if both hypotheses are tru This function is continuous on its domain, #9-oo, oo)#, so it is continuous on #[0, 4]#
Now, #f(x) =abs(x-3) = { (-x+3, ", if x < 3"), ( x-3, ", if x > 3") :}#

So, for the derivative, we have:

#f'(x) = { (-1, ", if x < 3"), ( 1, ", if x > 3") :}#
So #f# is not differentiable at #3#, so it is not differentiable on #(0, 0)#
This function does not satisfy the hypotheses on #[0,4]#.
In addition , since the derivative is always #-1# or #1# (where it exists),
and #(f(4)-f(0))/(4-0) = 1/2#
There is no #c in (0,4)# at which #f'(c) = (f(4)-f(0))/(4-0) #.
(There is no #c# that satisfies the conclusion.)
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Answer 2

To determine if the function ( f(x) = |x - 3| ) satisfies the hypotheses of the Mean Value Theorem on the interval ([0, 4]), we need to check if it meets two conditions:

  1. Continuity on the closed interval ([0, 4]).
  2. Differentiability on the open interval ((0, 4)).

Since ( |x - 3| ) is continuous and differentiable everywhere except at ( x = 3 ), it satisfies the first condition. To verify the second condition, we need to ensure that ( f'(x) ) exists for all ( x ) in ((0, 4)), except possibly at ( x = 3 ).

To find all numbers ( c ) on ([0, 4]) that satisfy the Mean Value Theorem, we need to find where the derivative of ( f(x) ) equals the average rate of change of ( f(x) ) over the interval ([0, 4]).

  1. Calculate the average rate of change of ( f(x) ) over ([0, 4]):

[ \frac{f(4) - f(0)}{4 - 0} = \frac{|4 - 3| - |0 - 3|}{4} = \frac{1 - 3}{4} = -\frac{1}{4} ]

  1. Find ( f'(x) ):

[ f'(x) = \frac{x - 3}{|x - 3|} ]

  1. Set ( f'(c) = -\frac{1}{4} ) and solve for ( c ):

[ \frac{c - 3}{|c - 3|} = -\frac{1}{4} ]

This equation holds true for ( c ) such that ( 0 < c < 3 ) and ( 3 < c < 4 ).

So, the numbers ( c ) satisfying the Mean Value Theorem on ([0, 4]) are in the intervals ( (0, 3) ) and ( (3, 4) ).

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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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