1. Summary of curve sketching

1. Summary of curve sketching Such that f’(x)=0 or f”(x) does not exist

(ii) x-intercept: if possible solve f(x)=0 (i) Horizontal Asymptotes

(i) Symmetry 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) Period: If f(x + p)=f(x) for all x in its Then f(x) is periodic with period p.

2(i): f(x)=2x3+5x2 - 4x Example Solution: All real numbers. f′(x)= 6x2 +10x -4 =2(3x2 +5x-2)=0

f′(-2)=12 f′(1/3)=-19/27

f″(x)= (-5/6, f(-5/6))=(-5/6, 5.65)

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

(i) Symmetry f(-x)=2(-x)3 +5(-x)2 – 4(-x) =-2x3 +5x2 + 4x No symmetry

f′(x)

f′(x) f″(x)

Example: Solution: f′(x)= f′(x)=0 ˂═˃

f″(x)= f″(x) f(x)

f(x) _ _ + f″(x)

There are no points of inflection since X=±1 are not in the domain of f(x). f(0)= y-intercept: x-intercept:

Horizontal

2(-∞)= 2(∞)= 2(∞)= 2(-∞)=

f(-x)= f(x)

Example 2(iii) f(x)=2sin(x)+sin2(x) Period: f(x+2π) =2sin(x+2π)+sin2(x+2π) =2sin(x)+sin2(x)=f(x)

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 or sin(x)=-1

f(π/2)=2+1 f(3π/2)=-2+1=-1

f″(x)= f″(x)=0 ˂=˃ sin(x)=-1, or sin(x)=1/2

f(π/6) f(π/6)

y-intercept: f(0)=2 sin(0)+ sin 2(0)=0 (0,0) is the y-intercept x-intercept: f(x)= 2 sin(x)+sin2 (x)=0 sin(x)(2+sin(x))=0 sin (x)=0 ═˃ (h) Asymptotes: No asymptotes.

Example: All real numbers. f′(x)= f′(x)=o =˃ f′(x) does not exist if x=0 ,

f(-1)=-(2)2/3

f″(x)=

y-intercept: f(0)=0 x1/3 (x+3)2/3 =0 x-intercept: Limx→∞ x1/3 (x+3)2/3 =∞ Limx→-∞ x1/3 (x+3)2/3 =-∞

f(-x)≠ f(x) f(-x)≠- f(x)

1. Summary of curve sketching