Given the function #f(x)=(-x^2+9)/(4x)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,3] and find the c?

Answer 1

The only hypothesis required by the mean value theorem is that #f(x)# has to be continuous in the interval.

The only hypothesis required by the mean value theorem is that #f(x)# has to be continuous in the interval.

As:

#f(x) = (9-x^2)/(4x)#
is in fact continuous in #[1,3]# the hypothesis is satisfied and we have that the is a value #c in [1,3]# for which:
#f(c) = 1/2 int_1^3 (9-x^2)/(4x)dx#

We can calculate the definite integral as:

#int_1^3 (9-x^2)/(4x)dx = int_1^3 (9/(4x)-x/4)dx=[9/4lnx-x^2/8]_1^3= 9/4ln3-9/8+1/8=9/4ln3-1#

Then we can find #c# from:
#f(c) = (9-c^2)/(4c) = 9/4ln3-1#
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Answer 2

I will assume that you are referring to the Mean Value Theorem for Derivatives.

You determine whether it satisfies the hypotheses by determining whether #f(x) = (-x^2+9)/(4x)# is continuous on the interval #[1,3]# and differentiable on the interval #(1,3)#. (Those are the hypotheses of the Mean Value Theorem)
You find the #c# mentioned in the conclusion of the theorem by solving #f'(x) = (f(3)-f(1))/(3-1)# on the interval #(1,3)#.
#f# is continuous on its domain, which includes #[1,3]#
#f'(x) = -(x^2+9)/(4x^2)# which exists for all #x != 0# so it exists for all #x# in #(1,3)#

Therefore this function satisfies the hypotheses of the Mean Value Theorem on this interval.

To find #c# solve the equation #f'(x) = (f(3)-f(1))/(3-1)#. Discard any solutions outside #(1,3)#.
You should get #c = sqrt3#. (The solution, #-sqrt3# is not in the interval.)
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Answer 3

To determine if the function ( f(x) = \frac{{-x^2 + 9}}{{4x}} ) satisfies the hypotheses of the Mean Value Theorem on the interval ([1, 3]) and find ( c ), you need to check two conditions:

  1. Continuity: Verify if ( f(x) ) is continuous on the interval ([1, 3]).
  2. Differentiability: Confirm if ( f(x) ) is differentiable on the interval ((1, 3)).

Once you've confirmed both conditions, you can find ( c ) by applying the Mean Value Theorem, which states that if a function ( f ) is continuous on ([a, b]) and differentiable on ((a, b)), then there exists a ( c ) in ((a, b)) such that ( f'(c) = \frac{{f(b) - f(a)}}{{b - a}} ).

After verifying the hypotheses of the Mean Value Theorem, to find ( c ), calculate ( f(1) ) and ( f(3) ), then find the derivative ( f'(x) ). Finally, use the formula ( f'(c) = \frac{{f(3) - f(1)}}{{3 - 1}} ) to solve 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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