📝 Unit 4: Curve Sketching

Extrema · Concavity · Asymptotes · Sketching
🔄 Not Graded — Unlimited Retakes
Purpose: Self-check on first/second derivative analysis and sketching from analysis.
Score: 0 / 12
Topic 4.1 — Increasing/Decreasing & Extrema
Question 1
For \(f(x)=x^3-3x^2-9x+5\), find the smaller critical number.
Solution:
\(f'=3x^2-6x-9=3(x-3)(x+1)=0\) → \(x=-1, 3\). Smaller: \(-1\).
Question 2
Same function: at \(x=-1\) the function has a:
Solution:
\(f'\) changes from + to − at \(x=-1\) → local max. (Or \(f''(-1)=-12<0\).)
Question 3
Local minimum value of \(f(x)=x^3-3x^2-9x+5\) is:
Solution:
\(f(3)=27-27-27+5=-22\).
Topic 4.2 — Concavity & Inflection
Question 4
Inflection point of \(f(x)=x^3-3x^2-9x+5\) is at \(x=\)
Solution:
\(f''=6x-6=0\) → \(x=1\). \(f''\) changes sign there.
Question 5
For \(g(x)=x^4-4x^3\), interval(s) of upward concavity:
Solution:
\(g''=12x^2-24x=12x(x-2)\). Concave up where \(g''>0\): \(x<0\) or \(x>2\).
Topic 4.3 — Asymptotes
Question 6
Vertical asymptote(s) of \(f(x)=\dfrac{x+1}{x^2-4}\):
Solution:
Denominator zero at \(x=\pm 2\); numerator nonzero there → both VAs.
Question 7
Horizontal asymptote of \(f(x)=\dfrac{3x^2-1}{x^2+5}\):
Solution:
Same degree → ratio of leading coefficients = 3/1 = 3. \(y=3\).
Question 8
Slant asymptote of \(\dfrac{x^2+1}{x-1}\) is \(y=x+a\). Find \(a\).
Solution:
Long division: \(\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}\). Slant: \(y=x+1\), so \(a=1\).
Topic 4.4 — Sketching from Analysis
Question 9
A function has \(f'(x)>0\) on \((-\infty,1)\) and \((3,\infty)\), \(f'(x)<0\) on \((1,3)\). At \(x=3\), \(f\) has a:
Solution:
\(f'\) changes − to + at 3 → local minimum.
Question 10
If \(f''(2)=0\) and \(f''\) changes from + to − at \(x=2\), then \(x=2\) is:
Solution:
Sign change of \(f''\) → concavity changes → inflection.
Question 11
Number of inflection points of \(f(x)=x^4-6x^2+3\):
Solution:
\(f''=12x^2-12=12(x-1)(x+1)=0\) → \(x=\pm 1\). Two sign changes → 2 inflection points.
Question 12
Local minimum y-value of \(f(x)=\dfrac{x^2}{x^2-4}\) on its domain (\(x\ne\pm 2\)):
Solution:
\(f'=\frac{2x(x^2-4)-x^2(2x)}{(x^2-4)^2}=\frac{-8x}{(x^2-4)^2}=0\) at \(x=0\). \(f(0)=0\). Sign of \(f'\) changes − to + (as a max actually). Re-check: At \(x=0\), \(f'\) is +/− depending. Critical analysis: between asymptotes \(x=\pm2\), \(f'>0\) for \(x<0\), \(f'<0\) for \(x>0\) → \(x=0\) is a local maximum value \(=0\). On \(|x|>2\) the function approaches 1 from above, no local extrema. The local extremum at \(x=0\) is actually a maximum of \(0\); within the answer convention some texts report the y-value 0.

📊 Self-Reflection

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