# How do you determine all values of c that satisfy the mean value theorem on the interval [1,9] for #f(x)=x^-4#?

As f(x) is strictly decreasing in the interval [1,9], there can be only one point in the interval satisfying the theorem.

Evaluate the integral:

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To determine all values of ( c ) that satisfy the Mean Value Theorem on the interval ([1, 9]) for ( f(x) = x^{-4} ), you first find the average rate of change of ( f(x) ) over that interval. Then, you find the derivative of ( f(x) ) and evaluate it at ( c ), the value guaranteed by the Mean Value Theorem.

The Mean Value Theorem states that if a function ( f(x) ) is continuous on the closed interval ([a, b]) and differentiable on the open interval ((a, b)), then there exists at least one number ( c ) in ((a, b)) such that ( f'(c) = \frac{f(b) - f(a)}{b - a} ).

In this case, the function is ( f(x) = x^{-4} ), and its derivative is ( f'(x) = -4x^{-5} ).

The average rate of change of ( f(x) ) over ([1, 9]) is:

[ \frac{f(9) - f(1)}{9 - 1} = \frac{1/6561 - 1}{8} = -\frac{3280}{6561} ]

So, to find the value of ( c ), we set:

[ f'(c) = -4c^{-5} = -\frac{3280}{6561} ]

Solving for ( c ), we get:

[ c^5 = \frac{6561}{3280} ]

[ c = \sqrt[5]{\frac{6561}{3280}} ]

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