Calculus BC ยท Unit 5
Find where the derivative is zero or undefined, and know when absolute extrema must exist.
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A critical point is in the domain
An interior point \(c\) is a critical point when \(f'(c) = 0\) or \(f'(c)\) does not exist.
A critical point is not automatically an extremum
\(f(x) = x^3\) has a critical point at \(x = 0\), and the sign of \(f'\) does not change there. The point is not a local maximum or a local minimum.
The Extreme Value Theorem
If \(f\) is continuous on a closed interval \([a, b]\), then \(f\) has an absolute maximum and an absolute minimum on that interval.
Where those absolute values can sit
On a closed interval, an absolute extremum occurs at a critical point in the interval or at an endpoint.
Set the derivative equal to 0, or find where it fails.
\(f'(x) = 2x - 6\)
\(2x - 6 = 0\)
\(x = 3\)
Answer: \(x = 3\)
\(f'(x) = 3x^2\)
\(3x^2 = 0\)
\(x = 0\)
\(f'(x) \ge 0\) on both sides, so this critical point is not a local extremum.
Answer: \(x = 0\)
\(f'(x) = \dfrac{2}{3} x^{-1/3} = \dfrac{2}{3\sqrt[3]{x}}\)
The derivative is never 0.
The derivative is undefined at \(x = 0\), and 0 is in the domain.
\(x = 0\)
Answer: \(x = 0\)
For \(x > 2\), \(f'(x) = 1\).
For \(x < 2\), \(f'(x) = -1\).
The derivative does not exist at \(x = 2\).
Answer: \(x = 2\)
The interval is closed and \(f\) is continuous on it.
f has an absolute maximum value and an absolute minimum value on \([0, 5]\).
Answer: an absolute max and an absolute min on \([0, 5]\)
The interval \((0, 1]\) is not closed.
The Extreme Value Theorem does not apply.
Answer: no
A cubic, a logarithm, or a vertical tangent.
\(f'(x) = 3x^2 - 3\)
\(3(x^2 - 1) = 0\)
\(x = -1\) and \(x = 1\)
Answer: \(x = -1\) and \(x = 1\)
\(f'(x) = \dfrac{2x}{x^2 + 1}\)
The denominator is never 0.
\(2x = 0\)
\(x = 0\)
Answer: \(x = 0\)
\(f'(x) = 1 - \dfrac{1}{x^2}\)
\(x^2 = 1\)
On \((0, \infty)\), \(x = 1\).
Answer: \(x = 1\)
\(f'(x) = \dfrac{1}{3} x^{-2/3}\)
The derivative is never 0, and it is undefined at \(x = 0\).
\(f'(x) > 0\) on both sides of 0, so f is increasing through the critical point.
There is a critical point at \(x = 0\), and it is not a local extremum.
Answer: critical at 0; not a local extremum
\(f\) is continuous on the closed interval \([1, 3]\).
The theorem applies. An absolute maximum and an absolute minimum exist on \([1, 3]\).
Answer: yes
Absolute extrema of a continuous function on a closed interval occur at critical points or endpoints.
Answer: A. critical points and endpoints
A product, an exponential, and a list of candidates.
Product rule.
\(f'(x) = 2(x - 1)(x + 2) + (x - 1)^2\)
\(f'(x) = (x - 1)\bigl[2(x + 2) + (x - 1)\bigr]\)
\(f'(x) = (x - 1)(3x + 3)\)
\(f'(x) = 3(x - 1)(x + 1)\)
\(x = 1\) and \(x = -1\)
Answer: \(x = -1\) and \(x = 1\)
\(f'(x) = e^{-x} - x e^{-x}\)
\(f'(x) = e^{-x}(1 - x)\)
\(e^{-x}\) is never 0.
\(x = 1\)
Answer: \(x = 1\)
\(f'(x) = \dfrac{2}{3} x^{-1/3}\) is undefined at \(x = 0\) and never 0.
The candidates are the critical point and the endpoints.
\(x = -1\), \(x = 0\), and \(x = 8\)
Answer: \(x = -1, 0, 8\)
On \([1, 4]\), \(f\) is continuous on a closed bounded interval.
On \((0, 1]\), the interval is not closed.
Answer: yes on \([1, 4]\); no on \((0, 1]\)
\(f'(x) = 1 - \dfrac{4}{x^2}\)
\(x^2 = 4\)
On \(x > 0\), \(x = 2\).
Answer: A. \(2\)
\(f'(3) = 0\), so \(x = 3\) is a critical point.
\((x - 3)^2\) does not change sign.
A critical point with no sign change is not a local extremum.
Answer: critical; not a local extremum
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