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Unformatted text preview: Homework # 6 Due: Never 1 New Material 1. Use the guidelines of this section to sketch the curve. (a) f ( x ) = 20 x 3 3 x 4 i. Domain f is a polynomial, so its domain is all real numbers. ii. Intercepts The yintercept is f (0) = 0. The xintercepts are where f ( x ) = 0, so 20 x 3 3 x 4 = 0. Factor out an x 3 and get x 3 (20 3 x ) = 0, so x = 0 , 20 / 3 iii. Symmetry There are no trig functions, so its not periodic. f ( x ) = 20( x ) 3 3( x ) 4 = 20 x 3 3 x 4 which is neither f ( x ) nor f ( x ), so f is neither even nor odd. iv. Asymptotes. Its not undefined anywhere, so there wont be any vertical asymptotes. lim x 20 x 3 3 x 4 = lim x x 3 (20 3 x ) = Using the same method, you can see that lim x  f ( x ) = and so there are no horizontal asymptotes. v. Intervals of Increase and Decrease f ( x ) = 60 x 2 12 x 3 . To look for intervals of increase and decrease, we first find the critical numbers. f ( x ) is a polynomial, so its defined everywhere. The only critical numbers are where f ( x ) = 0. 60 x 2 12 x 3 = 12 x 2 (5 x ), so f ( x ) = 0 when x = 0 , 5. This gives us 3 intervals: x f ( x ) x < + < x < 5 + 5 < x Increasing: ( , 0), (0 , 5) Decreasing: (5 , ) vi. Minima and Maxima At 0, the function doesnt change direction, so its neither a local max nor a min. At 5, the function changes from increasing to decreasing, so its a maximum. Local Maximum: x = 5 Local Minimum: None vii. Intervals of concavity and inflection points. f 00 ( x ) = 120 x 36 x 2 . Inflection points can be where this is zero or undefined. Its never undefined, so were only looking for f 00 ( x ) = 0. 120 x 36 x 2 = 0 so 12 x (10 3 x ) = 0. Thus possible inflection points are x = 0 , 10 / 3 and we have 3 intervals to check. x f 00 ( x ) x < < x < 10 / 3 + 10 / 3 < x CU: (0 , 10 / 3) CD: ( , 0), (10 / 3 , ) Inflection points at x = 0 , 10 / 3 since the concavity changes at both points. viii. Graph7.552.5 2.5 5 7.51000750500250 250 500 (b) f ( x ) = x 2 x 2 9 i. Domain Its a rational function, so its domain is everywhere except where the denom inator is 0. Thus the domain is all numbers except x = 3 , 3. ii. Intercepts yintercept: f (0) = 9 = 0, so the yintercept is 0. xintercept: x 2 x 2 9 = 0, so x = 0 is the xintercept. iii. Symmetry f ( x ) = ( x ) 2 ( x ) 2 9 = x 2 x 2 9 = f ( x ) so the function is even. iv. Asymptotes The function is undefined at x = 3 , 3, but were only interested in the positive x values since f is even. Lets look at one of the onesided limits at x = 3. lim x 3 + x 2 x 2 9 = lim x 3 + 1 1 9 /x 2 = Its since the denominator goes to zero and the numerator goes to 1, so the whole thing is an infinite limit. Its positive because when x > 3, 1 9 /x 2 > 0....
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 Spring '08
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 Real Numbers, YIntercept

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