How do you verify that the function #f(x)=x/(x+6)# satisfies the hypotheses of The Mean Value Theorem on the given interval [0,1] and then find the number c that satisfy the conclusion of The Mean Value Theorem?

Answer 1
The MVT states that if f(x) is continuous in [a,b] (it obviously is) and derivable in (a,b) (it obviously is too), then #exists# at least one #c in (a,b) : f(b)-f(a)=f'(c)(b-a)#
Notice the theorem doesn't give you the number of #c#s nor their values.

So we find them out:

#f(0)-f(1)=f'(c)(0-1) => f'(c)=1/7#

i.e.

#1/7=((c+6) - c)/(c+6)^2 => (c+6)^2=42 => c_1=-6+sqrt(42), c_2=-6-sqrt(42)#
We notice #c_2<0#, so #c_2# is not a root to be considered for MVT, the only choice we have left is #c_1#, and MVT assures us #c_1 in (0,1)# without any kind of manual verification
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Answer 2

To verify that the function ( f(x) = \frac{x}{x + 6} ) satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval ([0,1]), you need to check two conditions:

  1. ( f(x) ) is continuous on ([0,1]).
  2. ( f(x) ) is differentiable on ((0,1)).

First, let's check if ( f(x) ) is continuous on ([0,1]). The function ( f(x) ) is a rational function, and it is continuous everywhere in its domain except where its denominator ( x + 6 ) equals zero. However, on the interval ([0,1]), ( x + 6 ) is never zero, so ( f(x) ) is continuous on ([0,1]).

Next, let's check if ( f(x) ) is differentiable on ((0,1)). ( f(x) ) is differentiable everywhere in its domain except where it's not defined or where its derivative doesn't exist. Since ( f(x) ) is defined and continuous on ([0,1]), it is also differentiable on ((0,1)).

Since both conditions are satisfied, we can apply the Mean Value Theorem. According to MVT, there exists a number ( c ) in ((0,1)) such that ( f'(c) ) equals the average rate of change of ( f(x) ) over the interval ([0,1]).

To find ( c ), first, compute ( f'(x) ). Then, find the average rate of change of ( f(x) ) over ([0,1]). Finally, solve ( f'(c) = ) average rate of change for ( c ).

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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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