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Published byJoshua Wilkerson Modified over 9 years ago
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1. Summary of curve sketching Such that f’(x)=0 or f”(x) does not exist
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(ii) x-intercept: if possible solve f(x)=0 (i) Horizontal Asymptotes
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If f(-x)=f(x), then f(x) is symmetric w.r.t. the y- axis. If f(-x)=-f(x), then f(x) is symmetric w.r.t. the origin. (i) Symmetry
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(i) Period: If f(x + p)=f(x) for all x in its Then f(x) is periodic with period p.
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Example Solution: All real numbers. 2(i) : f(x)=2x 3 +5x 2 - 4x f′(x)= 6x 2 +10x -4 =2(3x 2 +5x-2)=0
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f′(-2)=12 f′(1/3)=-19/27
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f″(x)= (-5/6, f(-5/6))=(-5/6, 5.65)
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y-intercept: f(0)=0 x-intercept: f(x)=0, when x=0, as x→∞, f(x)→∞ as x→-∞, f(x)→-∞ No Horizontal Asymptotes or Vertical Asymptotes
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(i) Symmetry f(-x)=2(-x) 3 +5(-x) 2 – 4(-x) =-2x 3 +5x 2 + 4x No symmetry
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f′(x)
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f″(x)
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Example: Solution: f′(x)= f′(x)=0 ˂ ═ ˃
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f″(x)f(x) f″(x)=
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f(x) f″(x) _ + _
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There are no points of inflection since X=±1 are not in the domain of f(x). y-intercept: f(0)= x-intercept:
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Horizontal
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2(-∞)= 2(∞)= 2(-∞)=
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f(-x)= f(x)
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Period: Example 2(iii) f(x)=2sin(x)+sin 2 (x) f(x+2π) =2sin(x+2π)+sin 2 (x+2π) =2sin(x)+sin 2 (x)=f(x)
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All real numbers. f′(x)=2 cos(x)+2 sin(x) cos(x) =2 cos(x)(1+sin(x)) f′(x)=0 ˂ = ˃ 2cos(x)=0 orsin(x)=-1
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f(π/2)=2+1 f(3π/2)=-2+1=-1
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f″(x)= f″(x)=0 ˂ = ˃ sin(x)=-1, orsin(x)=1/2
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f(π/6)
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No asymptotes. (h) Asymptotes: y-intercept: f(0)=2 sin(0)+ sin 2 (0)=0 (0,0) is the y-intercept x-intercept: f(x)= 2 sin(x)+sin 2 (x)=0 sin(x)(2+sin(x))=0 sin (x)=0 ═ ˃
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Example: All real numbers. f′(x)= f′(x)=o = ˃ f′(x) does not exist if x=0,
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f(-1)=-(2) 2/3
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f″(x)=
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y-intercept: f(0)=0 x-intercept: x 1/3 (x+3) 2/3 =0 Lim x→∞ x 1/3 (x+3) 2/3 =∞ Lim x→-∞ x 1/3 (x+3) 2/3 =-∞
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f(-x)≠ f(x) f(-x)≠- f(x)
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