Given the function #f(x)=x/(x+9)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,18] and find the c?

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) = x/(x+9)# , #a=1# and #b=18#.
H1 is true because this function is a rational function and rational functions are continuous on their domains. So #f# is continuouos except at #x=-9# which is not in #[1,18]#
H2 is true because #f'(x) = 9/(x+9)^2# exists for all #x# except #x=-9# which is not in #(1,18)#
The conclusion of MVT says there is a number #c# in #(a,b)# such that #f'(c) = (f(b)-f(a))/(b-a)#. To find the #c#, solve the equation. Discard any solutions outside the interval #(a,b)#.

So we need to solve

#9/(x+9)^2 = ((18/27)-(1/10))/17# on #(1,18)#.
I get #-9+3sqrt30#
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Answer 2

To determine if the function ( f(x) = \frac{x}{x+9} ) satisfies the hypotheses of the Mean Value Theorem on the interval ([1,18]), we need to check two conditions:

  1. The function ( f(x) ) must be continuous on the closed interval ([1,18]).
  2. The function ( f(x) ) must be differentiable on the open interval ((1,18)).

To check the first condition, we verify if ( f(x) ) is continuous at every point in the interval ([1,18]). Since ( f(x) ) is a rational function, it is continuous for all real numbers except where the denominator is zero. However, ( x+9 \neq 0 ) for all ( x ) in the interval ([1,18]), so ( f(x) ) is continuous on ([1,18]).

Next, we check the second condition by determining if ( f(x) ) is differentiable on the open interval ((1,18)). To find the derivative, ( f'(x) ), we use the quotient rule:

[ f'(x) = \frac{(x+9) - x(1)}{(x+9)^2} = \frac{9}{(x+9)^2} ]

Since ( f'(x) ) is a continuous function for all ( x ) except where the denominator is zero, which is never the case in the interval ([1,18]), ( f(x) ) is differentiable on ((1,18)).

Therefore, since ( f(x) ) is continuous on ([1,18]) and differentiable on ((1,18)), it satisfies the hypotheses of the Mean Value Theorem on the interval ([1,18]). To find the value of ( c ), we use the Mean Value Theorem formula:

[ f'(c) = \frac{f(18) - f(1)}{18 - 1} ]

Plugging in the values, we get:

[ f'(c) = \frac{\frac{18}{18+9} - \frac{1}{1+9}}{17} ] [ f'(c) = \frac{\frac{18}{27} - \frac{1}{10}}{17} ] [ f'(c) = \frac{\frac{2}{3} - \frac{1}{10}}{17} ] [ f'(c) = \frac{\frac{20}{30} - \frac{3}{30}}{17} ] [ f'(c) = \frac{\frac{17}{30}}{17} ] [ f'(c) = \frac{1}{30} ]

Therefore, ( f'(c) = \frac{1}{30} ) for some ( c ) in the interval ((1,18)).

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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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