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Applications of Differentiation

3.6 A Summary of Curve Sketching

Focus on combine derivative information, asymptotes and excluded inputs, end behavior.

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Ideas in this lesson

01Combine derivative informationWorked steps · five questions02Asymptotes and excluded inputsWorked steps · five questions03End behaviorWorked steps · five questions
Idea 1 of 3

Combine derivative information

Use first derivatives for extrema and second derivatives for concavity; neither replaces domain and intercept analysis.

Worked example 1

For f(x)=x3−12xf(x) = x^{3} - 12x, find the inflection-point xx-coordinate.
Coordinate graph-3-2-1123-20-101020xy
  1. f′(x)=3(x2−4)  ⟹  x=±2f'(x) = 3(x^{2} - 4) \implies x = \pm 2
  2. f′′(x)=6xf''(x) = 6x; concavity changes at x=0x = 0
  3. Local max⁡=(−2,16)\max = (-2, 16), local min⁡=(2,−16)\min = (2, -16), inflection =(0,0)= (0, 0)
  4. Intercepts x=0,± 12x = 0, \pm\,\sqrt{12}; f(x)→±∞f(x) \to \pm \infty as x→±∞x \to \pm \infty ; requested value =0= 0

Answer: 00

Worked example 2

For f(x)=x3−27xf(x) = x^{3} - 27x, find the local minimum xx-coordinate.
Coordinate graph-4-3-2-11234-80-60-40-2020406080xy
  1. f′(x)=3(x2−9)  ⟹  x=±3f'(x) = 3(x^{2} - 9) \implies x = \pm 3
  2. f′′(x)=6xf''(x) = 6x; concavity changes at x=0x = 0
  3. Local max⁡=(−3,54)\max = (-3, 54), local min⁡=(3,−54)\min = (3, -54), inflection =(0,0)= (0, 0)
  4. Intercepts x=0,± 27x = 0, \pm\,\sqrt{27}; f(x)→±∞f(x) \to \pm \infty as x→±∞x \to \pm \infty ; requested value =3= 3

Answer: 33

Idea 2 of 3

Asymptotes and excluded inputs

Distinguish noncanceled denominator zeros from removable holes before drawing branches.

Worked example 1

Find the vertical asymptote of f(x)=1x−2f(x) = \dfrac{1}{x - 2}.
  1. x−2=0  ⟹  x=2x - 2 = 0 \implies x = 2
  2. Numerator at x=2x = 2 is 1≠01 \ne 0
  3. Vertical asymptote: x=2x = 2

Answer: 22

Worked example 2

Find the vertical asymptote of f(x)=x+1x−3f(x) = \dfrac{x + 1}{x - 3}.
  1. x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3
  2. Numerator at x=3x = 3 is 4≠04 \ne 0
  3. Vertical asymptote: x=3x = 3

Answer: 33

Idea 3 of 3

End behavior

Use limits at infinity alongside extrema and inflection points to organize a complete sketch.

Worked example 1

Find the horizontal asymptote height of f(x)=2x2−1x2+2f(x) = \dfrac{2x^{2} - 1}{x^{2} + 2}.
  1. f(x)=2−1x21+2x2f(x) = \dfrac{2 - \dfrac{1}{x^{2}}}{1 + \dfrac{2}{x^{2}}}
  2. lim⁡x→±∞f(x)=21\displaystyle \lim\limits_{x\to \pm \infty } f(x) = \dfrac{2}{1}
  3. Horizontal asymptote: y=2y = 2

Answer: 22

Worked example 2

Find the horizontal asymptote height of f(x)=3x2−12x2+2f(x) = \dfrac{3x^{2} - 1}{2x^{2} + 2}.
  1. f(x)=3−1x22+2x2f(x) = \dfrac{3 - \dfrac{1}{x^{2}}}{2 + \dfrac{2}{x^{2}}}
  2. lim⁡x→±∞f(x)=32\displaystyle \lim\limits_{x\to \pm \infty } f(x) = \dfrac{3}{2}
  3. Horizontal asymptote: y=32≈1.5y = \dfrac{3}{2} \approx 1.5

Answer: 32≈1.5\dfrac{3}{2} \approx 1.5

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §3.6,p3.6, p. 206206. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.