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

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

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

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

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) =-2x 3 +5x 2 + 4x No symmetry

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). y-intercept: f(0)= x-intercept:

Horizontal

2(-∞)= 2(∞)= 2(-∞)=

f(-x)= f(x)

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)

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

f(π/2)=2+1 f(3π/2)=-2+1=-1

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

f(π/6)

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 ═ ˃

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 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 =-∞

f(-x)≠ f(x) f(-x)≠- f(x)