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Unformatted text preview: Explanation: ±or a function of the form F ( x ) = ( x-a )( x-b ) x-m we see that F ( a ) = F ( b ); in addition, since the denominator is zero only at x = m , F is continuous and di²erentiable on (-∞ , m ) [ ( m, ∞ ) . Thus Rolle’s Theorem can be applied to F on the interval [ a, b ] so long as m does not belong to ( a, b ). When f ( x ) = ( x + 6)( x-3) x + 9 , therefore, Rolle’s Theorem applies to f on the interval [-6 , 3]. Now, by the Quotient Rule, f ( x ) = (2 x + 3)( x + 9)-( x 2 + 3 x-18) ( x + 9) 2 = x 2 + 18 x + 45 ( x + 9) 2 . But x 2 + 18 x + 45 = ( x + 3)( x + 15) , so f (-3) = 0 = f (-15) ....
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.
- Spring '08