# Using the principle of the mean-value theorem on the indicated interval, how do you find all numbers c that satisfy the conclusion of the theorem #f(x) = x³ + 5x² - 2x - 5#; [-1, 2]?

The answer is

By the mean value theorem,

Also,

Therefore,

So,

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To find all numbers ( c ) that satisfy the conclusion of the Mean Value Theorem for the function ( f(x) = x^3 + 5x^2 - 2x - 5 ) on the interval ([-1, 2]), follow these steps:

- Calculate the derivative of ( f(x) ) to find ( f'(x) ).
- Evaluate ( f'(x) ) at each critical point within the interval ([-1, 2]) to determine the potential values of ( c ) that satisfy the Mean Value Theorem.

First, find the derivative of ( f(x) ):

[ f'(x) = 3x^2 + 10x - 2 ]

Next, find the critical points by setting ( f'(x) ) equal to zero and solving for ( x ):

[ 3x^2 + 10x - 2 = 0 ]

Solve for ( x ) using the quadratic formula:

[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ]

Plugging in ( a = 3 ), ( b = 10 ), and ( c = -2 ):

[ x = \frac{{-10 \pm \sqrt{{10^2 - 4(3)(-2)}}}}{{2(3)}} ] [ x = \frac{{-10 \pm \sqrt{{100 + 24}}}}{6} ] [ x = \frac{{-10 \pm \sqrt{124}}}{6} ] [ x = \frac{{-10 \pm 2\sqrt{31}}}{6} ] [ x = \frac{{-5 \pm \sqrt{31}}}{3} ]

Now, evaluate ( f'(x) ) at the endpoints and critical points within the interval ([-1, 2]):

- ( f'(-1) = 3(-1)^2 + 10(-1) - 2 = 1 )
- ( f'(2) = 3(2)^2 + 10(2) - 2 = 24 )
- ( f'\left(\frac{{-5 + \sqrt{31}}}{3}\right) ) (approximately -1.28)
- ( f'\left(\frac{{-5 - \sqrt{31}}}{3}\right) ) (approximately -8.72)

Since ( f'(-1) = 1 ) and ( f'(2) = 24 ), by the Mean Value Theorem, there exists at least one ( c ) in ((-1, 2)) such that ( f'(c) ) equals the average rate of change of ( f(x) ) on ([-1, 2]), which is:

[ \frac{{f(2) - f(-1)}}{{2 - (-1)}} = \frac{{f(2) - f(-1)}}{3} = \frac{{f(2) - f(-1)}}{3} = \frac{{27 - (-10)}}{3} = \frac{{37}}{3} ]

Thus, ( f'(c) = \frac{{37}}{3} ) for some ( c ) in ((-1, 2)). However, the exact value of ( c ) depends on solving ( f'(c) = \frac{{37}}{3} ), which requires numerical methods or additional analysis.

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