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Exam2_3 7 - f ′ x =(6 x(4 − x x 2 24 2 = 0 Thus the...

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Version 208 – Exam 2 – Radin – (58305) 7 Explanation: The vertical asymptotes of the graph of a rational function f , i.e. , a quotient f ( x ) = P ( x ) Q ( x ) of polynomial functions P ( x ) and Q ( x ), occur at solutions of Q ( x ) = 0, while the graph will have a horizontal aymptote y = c only when degree P ( x ) = degree Q ( x ) and lim x → ±∞ P ( x ) Q ( x ) = c . Now parenleftbigg x + 1 5 parenrightbigg ( x 5) = 1 5 (5 x + 1) ( x 5) = 1 5 ( 5 x 2 24 x 5 ) . Thus Q ( x ) must contain the factor 5 x 2 24 x 5 . On the other hand, lim x → ±∞ P ( x ) Q ( x ) = 5 . Consequently, of the possible choices for P ( x ) /Q ( x ), only the graph of f ( x ) = 25 x 2 2 x + 5 5 x 2 24 x 5 has vertical asymptotes x = 1 5 , 5 and hori- zontal asymptote y = 5. 015 10.0 points Which one of the following properties does f ( x ) = x + 1 x 2 + 24 have? 1. local max at x = 4 2. local min at x = 4 3. local max at x = 6 4. local min at x = 6 5. local min at x = 4 6. local max at x = 4 correct Explanation: By the Quotient rule, f ( x ) = x 2 + 24 2 x ( x + 1) ( x 2 + 24) 2 = 24 2 x x 2 ( x 2 + 24) 2 . The critical points of f occur when f ( x ) = 0, i.e.
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Unformatted text preview: f ′ ( x ) = (6 + x )(4 − x ) ( x 2 + 24) 2 = 0 . Thus the critical points of f are x = − 6 and x = 4. To classify these critical points we use the First Derivative Test. But the sign of f ′ de-pends only on the numerator, so it is enough, therefore, to look only at a sign chart for (6 + x )(4 − x ): − 6 4 − − + From this it follows that f is decreasing on ( −∞ , − 6), increasing on ( − 6 , 4), and de-creasing on (4 , ∞ ). Consequently, f has a local maximum at x = 4 . keywords: local maximum, local minimum, critical point, quotient rule, First Derivative Test, rational function 016 10.0 points Find the interval(s) where f ( x ) = 2 3 x 4 − 1 3 x 3 − 3 x 2 − 5 x + 1...
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