Chapter 26: Curve Sketching and Oblique Asymptotes

Curve Sketching

Recall

Critical points \leftrightarrow  f1(x)=0f^{1}(x)=0 or undef

Increasing \rightarrow f1(x)>0f^{1}(x)>0

Decreasing \rightarrow f1(x)<0f^{1}(x)<0

Concave up f11(x)>0\rightarrow f^{11}(x)>0

Concave down f11(x)<0\rightarrow f^{11}(x)<0

Point of inflection\rightarrow  1) Tangent line

2) Concavity change


The Curve Sketching Algorithm

  1. Identify the domain, symmetries, intercepts
  2. Calculate f1(x)f^{1}(x) and f11(x)f^{11}(x)
  3. Find the critical points and their behaviour
  4. Increasing/decreasing
  5. Concavity and points of inflection
  6. Asymptoles
  7. Draw the curve

example

Sketch f(x)=x33xf(x)=x^{3}-3x

1) Domain =(δ,δ)=(-\delta ,\delta )

y-intercept

f(0)=0f(0)=0

x-intercept

0=x33x0=x^{3}-3x

x=0,3,3\Rightarrow x=0,-\sqrt{3}, \sqrt{3}

odd symmetry

f(x)=(x3)3(x)f(-x)=(-x^{3})-3(-x)

=x3+3x=f(x)=-x^{3}+3x=-f(x)

2) f1(x)=3x23f^{1}(x)=3x^{2}-3

f11(x)=6xf^{11}(x)=6x

3) If f1(x)=0f^{1}(x)=0 then 3x23=3(x21)=03x^{2}-3=3(x^{2}-1)=0

\rightarrow x=1x=-1 or x=+1x=+1

4) Interval f1(x)f^{1}(x) sign behaviour

5) f11(x)=6xf^{11}(x)=6x \rightarrow if x>0x>0 concave up

if x<0x<0 concave down

6) limxδf(x)limxδx(x23)=δ\lim_{x\rightarrow \delta } f(x)\lim_{x\rightarrow \delta }x(x^{2}-3)=\delta

limxδf(x)limxδx(x23)=δ\lim_{x\rightarrow -\delta } f(x)\lim_{x\rightarrow -\delta }x(x^{2}-3)=-\delta 

7)

Example

Sketch f(x)=(x+1)21+x2f(x) = \frac{(x+1)^{2}}{1+x^{2}}

Assume f1(x)=2(1x)2(1+x2)2f^{1}(x) = \frac{2(1-x)^{2}}{(1+x^{2})^{2}}

f11(x)=4x(x23)(1+x2)3f^{11}(x) = \frac{4x(x^{2}-3)}{(1+x^{2})^{3}}

1) Domain (δ,δ)(-\delta ,\delta )

f(x)=(x+1)21+(x)2=(1x)21+x2f(x)f(-x)=\frac{(-x+1)^{2}}{1+(-x)^{2}}=\frac{(1-x)^{2}}{1+x^{2}}\neq f(x)  No symmetry

y-intercept

f(0)=11=1f(0)=\frac{1}{1}=1

x-intercept

(x+1)21+x2=0x=1\frac{(x+1)^{2}}{1+x^{2}}=0\Rightarrow x=-1

2) f1(x)=2(x+1)(1+x2)(x+1)2(2x)(1+x2)2=2(1x2)(1+x2)2f^{1}(x)=\frac{2(x+1)(1+x^{2})-(x+1)^{2}(2x)}{(1+x^{2})^{2}}=\frac{2(1-x^{2})}{(1+x^{2})^{2}}

f11(x)=4x(x23)(1+x2)3f^{11}(x)=\frac{4x(x^{2}-3)}{(1+x^{2})^{3}}

3) f1(x)=0x=1orx=1f^{1}(x)=0 \rightarrow x=1 or x=-1

4) Interval sign f1(x)f^{1}(x) sign behaviour

5) f11(x)=4x(x23)(1+x2)3f^{11}(x)=\frac{4x(x^{2}-3)}{(1+x^{2})^{3}}

6) limxδf(x)=limxδx2+2x+11+x2=1\lim_{x\rightarrow \delta }f(x)=\lim_{x\rightarrow \delta }\frac{x^{2}+2x+1}{1+x^{2}}=1

limxδf(x)=1\lim_{x\rightarrow \delta }f(x)=1

Oblique Asymptotes

Define

f(x) has an OBLIQUE ASYMPTOTE

y=mx+by=mx+b

limxδ[f(x)(mx+b)]=0\lim_{x\rightarrow \delta }\left [ f(x)-(mx+b) \right ]=0 or

limxδ[f(x)(mx+b)]=0\lim_{x\rightarrow -\delta }\left [ f(x)-(mx+b) \right ]=0

example

The function f(x)=3x+1xf(x)=3x+\frac{1}{x}  has y=3x+0y=3x+0 as an oblique asymptote

limxδ[f(x)(3x+0)]=limxδ[3(x)+1x3x]\lim_{x\rightarrow \delta }\left [ f(x)-(3x+0) \right ]=\lim_{x\rightarrow \delta }\left [ 3(x)+\frac{1}{x}-3x \right ]

== limxδ[1x]=0\lim_{x\rightarrow \delta }\left [ \frac{1}{x} \right ]=0

example

Find the oblique asymptote of

g(x)=2x2+x+13x+1g(x)=\frac{2x^{2}+x+1}{3x+1}

g(x)(mx+b)=2x2+x+13x+1(mx+b)(3x+1)3x+1g(x)-(mx+b)=\frac{2x^{2}+x+1}{3x+1}-\frac{(mx+b)(3x+1)}{3x+1}

=2x2+x+1(3mx2+mx+3bx+b)3x+1=\frac{2x^{2}+x+1-(3mx^{2}+mx+3bx+b)}{3x+1}

=(23m)x2+(1m3b)x+(1b)3x+1=\frac{(2-3m)x^{2}+(1-m-3b)x+(1-b)}{3x+1}

We need 23m=02-3m=0  and 1m3b=01-m-3b=0

Thus, m=23m=\frac{2}{3} and b=19b=\frac{1}{9}

y=23x+19y=\frac{2}{3}x+\frac{1}{9}



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