Math 234 February 28. I.Find all vertical and horizontal asymptotes of the graph of the given function.
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1 Math 234 February 28 I.Find all vertical and horizontal asymptotes of the graph of the given function.. f(x) = /(x 3) x 3 = 0 when x = 3 Vertical Asymptotes: x = 3 H.A.: /(x 3) = 0 /(x 3) = 0 Horizontal asymptotes: y = 0 2. f(x) = (3x )/(x + 2) x + 2 = 0 when x = 2 Vertical Asymptotes: x = 2 H.A.: (3x )/(x + 2) = 3 (3x )/(x + 2) = 3 Horizontal Asymptotes: y = 3 3. f(x) = (3 2x)/(4 x) 4 x = 0 when x = 4 Vertical Asymptotes: x = 4 H.A.:
2 (3 2x)/(4 x) = 2 (3 2x)/(4 x) = 2 Horizontal Asymptotes: y = 2 4. f(x) = (x 2 + 3)/(x 2 + 4) x = 0 when x = 2, 2 Vertical Asymptotes: x = 2, 2 H.A.: (x 2 + 3)/(x 2 + 4) = (x 2 + 3)/(x 2 + 4) = Horizontal Asymptotes: y = 5. f(x) = (3x 2 + 3)/(4x 2 4) 4x 2 4 = 0 when x =, Vertical Asymptotes: x =, H.A.: (3x 2 + 3)/(4x 2 4) = 3/4 (3x 2 + 3)/(4x 2 4) = 3/4 Horizontal Asymptotes: y = 3/4 6. f(x) = (3x + 4)/x 2 x 2 = 0 when x = 0 Vertical Asymptotes: x = 0 H.A.: 2
3 (3x + 4)/x 2 = 0 (3x + 4)/x 2 = 0 Horizontal Asymptotes: y = 0 7. f(x) = (x 3 + 3x + 5)/(6x + 2) 6x + 2 = 0 when x = /3 Vertical Asymptotes: x = /3 H.A.: (x 3 + 3x + 5)/(6x + 2) = (x 3 + 3x + 5)/(6x + 2) = Horizontal Asymptotes: none 8. f(x) = (x x 2 + 3x + 5)/(2x 3 + 5x 2 + 6x + 2) 2x 3 + 5x 2 + 6x + 2 = 0 when x = /2 Vertical Asymptotes: x = /2 H.A.: (x x 2 + 3x + 5)/(2x 3 + 5x 2 + 6x + 2) = /2 (x x 2 + 3x + 5)/(2x 3 + 5x 2 + 6x + 2) = /2 Horizontal Asymptotes: y = /2 9. f(x) = 2/x /(3x ) 2/x = 0 when x = 0 /(3x ) = 0 when x = /3 Vertical Asymptotes: x = 0, /3 3
4 H.A.: 2 x 3x = 2(3x ) x x(3x ) = 5x (3x 2 x) = 0 2 x 3x = 5x (3x 2 x) = 0 Horizontal Asymptotes: y = 0 0. g(x) = 3x/( x 2 9) V.A. x2 9 = 0 when x = 3, 3 Vertical Asymptotes: x = 3, 3 H.A. 3x ( x 2 9) = 3x ( x 2 9) = 3x ( x 2 9) 3x ( x 2 9) Horizontal Asymptotes: y = 3, 3 (/ x 2 ) (/ x 2 ) = 3 = 3 9x 2 (/ x 2 ) (/ x 2 ) = 3x x 9x 2 = 3 9x 2 = 3. f(x) = x 2 / x 4 + V.A. x4 + > 0. Therefore there are no vertical asymptotes. Vertical Asymptotes: none H.A. x 2 x4 + = x 2 x4 + x 4 x 4 = = + x 4 4
5 x 2 x4 + = x 2 x4 + x 4 x 4 = x 2 x 2 + x 4 = + x 4 = Horizontal Asymptotes: y = II. Sketch the graph of the given function.. f(x) = /(x 3) Domain: x 3 y-intercept: f(0) = /3 x-intercept: none Vertical Asymptotes: x = 3 Horizontal Asymptotes: y = 0 f(x) = /(x 3) 2 Critical numbers: x = 3 f (x) < 0 for x/neq3 f(x) is decreasing: (, 3) (3, ) There are no extreme. f (x) = 2/(x 3) 3 f (x) < 0 for x < 3 and f (x) > 0 for x > 3 f(x) is concave down: (, 3) f(x) is concave up: (3, ) Since f(x) does not exist at x = 3 there is no inflection point to plot. The graph looks like: 5
6 f(x) = x 5 5x Domain: all real numbers y-intercept: f(0) = 93 x-intercept: f(x) = 0 when x =.9502, , (You have to use a calculator for this one) Vertical Asymptotes: none Horizontal Asymptotes: none f (x) = 5x 4 20x 3 = 5x 3 (x 4) Critical numbers: x = 0, 4 f (x) < 0 when 0 < x < 4 and f (x) > 0 when x < 0 or x > 4. f (x) is increasing: x < 0 or x > 4 f (x) is decreasing: 0 < x < 4 There is a relative max at (0, 93) and a relative min (4, 63). f (x) = 20x 3 60x 2 = 20x 2 (x 3) f (x) < 0 when x < 3 and f (x) > 0 when x > 3. f(x) is concave down: (, 3) f(x) is concave up: (3, ) There is an inflection point at (3, 69) The graph looks like: 6
7 f(x) = 3x 4 4x Domain: all real numbers y-intercept: f(0) = 3 x-intercept: none Vertical Asymptotes: none Horizontal Asymptotes: none f (x) = 2x 3 8x = 4x(3x 2 2) Critical numbers: x = 2/3, 0, 2/3 f (x) < 0 when x < 2/3 or 0 < x < 2/3; and f (x) > 0 when 2/3 < x < 0 or x > 2/3 f (x) is increasing: 2/3 < x < 0 or x > 2/3 f (x) is decreasing: x < 2/3 or 0 < x < 2/3 There is a relative maximum at (0, 3) and relative minimums at ( 2/3, 5/3) and ( 2/3, 5/3) f (x) = 36x 2 8 f (x) < 0 when 2/9 < x < 2/9 and f (x) > 0 when x < 2/9 or x > 2/9. f(x) is concave down: ( 2/9, 2/9) f(x) is concave up: (, 2/9) ( 2/9, ) There are inflection points at ( 2/9, 6/27) and ( 2/9, 6/27) The graph looks like: 7
8 f(x) = x 3 3x 4 Domain: all real numbers y-intercept: f(0) = 0 x-intercept:f(x) = 0 when x = 0, /3 Vertical Asymptotes: none Horizontal Asymptotes: none f (x) = 3x 2 2x 3 = 3x 2 ( 4x) Critical numbers: x = 0, /4 f (x) > 0 when x < /4 and f (x) < 0 when x > /4 f (x) is increasing: x < /4 f (x) is decreasing: x > /4 There is a relative maximum at (/4, /256) f (x) = 6x 36x 2 = 6x( 6x) f (x) > 0 when x < 0 or x > /6 and f (x) < 0 when 0 < x < /6 f(x) is concave down: (0, /6) f(x) is concave up: (, 0) (/6, ) There are inflection points at (0, 0) and (/6, /432) The graph looks like: 8
9 f(x) = /(2x + 3) Domain: x 3/2 y-intercept: f(0) = /3 x-intercept: none Vertical Asymptotes: x = 3/2 H.A. 2x + 3 = 0 2x + 3 = 0 Horizontal Asymptotes: y = 0 f (x) = 2/(2x + 3) 2 Critical numbers: x = 3/2 f(x) < 0 for x 3/2 f (x) is decreasing: x 3/2 There are no extreme. f (x) = 8/(2x + 3) 3 f (x) > 0 for x > 3/2 and f (x) < 0 for x < 3/2 f(x) is concave down: (, 3/2) f(x) is concave up: ( 3/2, ) Since f(x) does not exist at x = 3/2 there is no inflection point to plot. 9
10 The graph looks like: f(x) = x 2 /(x + 2) Domain: x 2 y-intercept: f(0) = 0 x-intercept: f(x) = 0 when x = 0 Vertical Asymptotes: x = 2 H.A. Horizontal Asymptotes: none f (x) = x 2 x + 2 = x 2 x + 2 = 2x(x + 2) x2 = x2 + 4x x(x + 4) = (x + 2) 2 (x + 2) 2 (x + 2) 2 Critical numbers: x = 4, 2, 0 f (x) < 0 when 4 < x < 2 or 2 < x < 0 f (x) > 0 when x < 4 or x > 0 f (x) is increasing: (, 4) (0, ) f (x) is decreasing: ( 4, 0) There is a relative maximum at ( 4, 8) and a relative minimum at 0
11 (0, 0) f (x) = (2x+4)(x+2)2 (x 2 +4x)(2(x+2)) (x+2) 4 = (x+2)[(2x+4)(x+2) 2x2 8x] (x+2) 4 = (x+2)[2x2 +8x+8 2x 2 8x] (x+2) 4 = (x+2)8 (x+2) 4 f (x) < 0 for x < 2 and f (x) > 0 for x > 2 f(x) concave down: x < 2 f(x) concave up: x > 2 Since f(x) does not exist at x = 2 there is no inflection point to plot. The graph looks like: f(x) = /(x 2 9) Domain: x 3, 3 y-intercept: f(0) = /9 x-intercept: None Vertical Asymptotes: x = 3, 3 H.A. x 2 9 = 0 x 2 9 = 0
12 Horizontal Asymptotes: y = 0 f 2x (x) = (x 2 9) 2 Critical numbers: x = 3, 0, 3 f (x) < 0 when x < 3 or 3 < x < 0; and f (x) > 0 when 0 < x < 3 or x > 3. f (x) is increasing: (, 3) ( 3, 0) f (x) is decreasing: (0, 3) (3, ) There is a relative maximum at (0, /9) f (x) = 2(x2 9) 2 2x2(x 2 9)2x (x 2 9) 4 = (x2 9)(2(x 2 9) 8x 2 ) (x 2 9) 4 = 6x2 8 (x 2 9) 3 = (6)( x2 3) (x 2 9) 3 f (x) < 0 when 3 < x < 3 ; and f (x) > 0 when x < 3 or x > 3. f(x) concave down: ( 3, 3) f(x) concave up: (, 3) (3, ) Since f(x) does not exist at x = 3 or x = 3 there is no inflection point to plot. The graph looks like: f(x) = / x 2 2
13 Domain: (, ) y-intercept: f(0) = x-intercept: None Vertical Asymptotes: x =, H.A. Horizontal Asymptotes: y = 0 f (x) x 2 = 0 x 2 = 0 = 0( x 2 ) ( 2 ( x2 ) /2 ( 2x)) ( x 2 ) 2 = x( x2 ) /2 ( x 2 ) 2 = x ( x 2 ) 3/2 Critical numbers: x = 0 and x < or x > (f (x) does not exist) f (x) > 0 for 0 < x < and f (x) < 0 for < x < 0 f (x) is increasing: (0, ) f (x) is decreasing: (, 0) There is a relative minimum at (0, ) f (x) = ( x2 ) 3/2 (3/2)( x 2 ) /2 ( 2x) (( x 2 ) 3/2 ) 2 = ( x2 ) 3/2 +3( x 2 ) /2 (( x 2 ) 3/2 ) 2 = ( x2 ) /2 (( x 2 )+3) ( x 2 ) 3 for < x < = (4 x2 ) ( x 2 ) 5/2 for < x < f (x) > 0 for < x < f(x) is concave up for the domain of f(x). There are no inflection points. The graph looks like: 3
14 f(x) = (x 2 9)/(x 2 + ) Domain: all real numbers y-intercept: f(0) = 9 x-intercept: f(x) = 0 when x = 3, 3 Vertical Asymptotes: none H.A. x 2 9 x 2 + = x 2 9 x 2 + = Horizontal Asymptotes: y = f (x) = 2x(x2 +) (x 2 9)(2x) (x 2 +) 2 = 2x+8x (x 2 +) 2 = 20x (x 2 +) 2 Critical numbers: x = 0 f (x) < 0 when x < 0 and f (x) > 0 when x > 0 f (x) is increasing: (0, ) f (x) is decreasing: (, 0) There is a relative minimum at (0, 9) 4
15 f (x) = 20((x2 +) 2 ) 20x2(x 2 +)2x) (x 2 +) 4 = (x2 +)(20(x 2 +) 80x 2 ) (x 2 +) 4 = (x2 +)( 60x (x 2 +) 4 = (x2 +)(20)( 3x 2 (x 2 +) 4 f (x) > 0 when /3 < x < /3; and f (x) < 0 when x < /3 or x > /3 f(x) concave up: ( /3, /3) f(x) concave down: (, /3) ( /3, ) There are inflection points at ( /3, 3/2) and ( /3, 3/2) The graph looks like: III. Find the absolute maximum and absolute minimum of the given function on the specified interval.. f(x) = x 3 + x 2 + [ 3, 2] f(x) is continuous on [ 3, 2]. Therefore the function has the extreme value property. First we find the critical numbers. f (x) = 3x 2 + 2x = x(3x + 2) f (x) = 0 when x = 0, 2/3. Both values are in the given interval f (x) < 0 when 2/3 < x < 0 and f(x) > 0 when x < 2/3 or x > 0. 5
16 There is a relative max at ( 2/3, 3/27) and a relative min at (0, ) At the endpoints, f( 3) = 7 and f(2) = 3 The absolute maximum is 3 and the absolute minimum is f(x) = (x 2 4) 5 [ 3, 2] f(x) is continuous on [ 3, 2]. Therefore the function has the extreme value property. First we find the critical numbers. f (x) = 5(x 2 4) 4 (2x) f (x) = 0 when x = 2, 0, 2. These values are in the given interval. f (x) < 0 when x < 2 or 2 < x < 0; and f (x) > 0 when 0 < x < 2 or x > 2. There is a relative minimum at (0, 024). At the endpoints, f( 3) = 325 and f(2) = 0. The absolute maximum is 325 and the absolute minimum is f(x) = x 2 /(x ) [ 2, /2] f(x) is continuous on[ 2, /2]. Therefore the function has the extreme value property. First we find the critical numbers. f (x) = 2x(x ) x2 = x2 2x x(x 2) = (x ) 2 (x ) 2 (x ) 2 f (x) = 0 when x = 0, 2. Neither value is in the given interval, therefore we only have to check the endpoints. At the endpoints, f( 2) = 4/3 and f( /2) = /6. The absolute maximum is -/6 and the absolute minimum is -4/3. 6
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