Calculus BC · Unit 5
Test every critical point and both endpoints. The largest and smallest values win.
The candidates test
On a closed interval, evaluate \(f\) at every critical point in the interval and at both endpoints. The largest value is the absolute maximum. The smallest value is the absolute minimum.
A local maximum can lose to an endpoint
Compare the numbers. The winner is the absolute extremum on that interval, even if a local extremum sits in the middle.
Include a critical point where f' does not exist
A corner or a cusp inside the interval is a candidate, the same way a zero of \(f'\) is.
An open interval can have no absolute extremum
The candidates test is for a continuous function on a closed interval. On an open interval, a function can keep rising and never reach a largest value.
One critical point, or none. Compare it with the endpoints.
\(f'(x) = 2x = 0\) at \(x = 0\)
\(f(-1) = 1\)
\(f(0) = 0\)
\(f(2) = 4\)
Absolute maximum 4 at \(x = 2\). Absolute minimum 0 at \(x = 0\).
Answer: max 4 at \(x = 2\); min 0 at \(x = 0\)
\(f'(x) = -2x = 0\) at \(x = 0\), an endpoint.
\(f(0) = 4\)
\(f(3) = -5\)
Answer: max 4 at \(x = 0\); min \(-5\) at \(x = 3\)
\(f'(x) = 3x^2 = 0\) at \(x = 0\)
\(f(-1) = -1\)
\(f(0) = 0\)
\(f(2) = 8\)
Answer: max 8 at \(x = 2\); min \(-1\) at \(x = -1\)
\(f'(x) = e^x > 0\), so there is no critical point.
f is increasing, so the minimum is at the left endpoint and the maximum is at the right endpoint.
\(f(0) = 1\) and \(f(1) = e\)
Answer: max \(e\) at \(x = 1\); min 1 at \(x = 0\)
\(f'(x) = \cos x = 0\) at \(x = \dfrac{\pi}{2}\)
\(f(0) = 0\)
\(f(\pi/2) = 1\)
\(f(\pi) = 0\)
Answer: max 1 at \(x = \pi/2\); min 0 at both endpoints
The interval is open.
Values of \(f\) get arbitrarily close to 1 and never reach a largest value on \((0, 1)\).
Answer: the interval is open, so no absolute max
Two critical points. Evaluate every candidate.
\(f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)\)
Candidates: \(x = 0, 1, 3, 4\)
\(f(0) = 1\)
\(f(1) = 5\)
\(f(3) = 1\)
\(f(4) = 5\)
Absolute maximum 5 at \(x = 1\) and \(x = 4\). Absolute minimum 1 at \(x = 0\) and \(x = 3\).
Answer: max 5; min 1
\(f'(x) = 1 - \dfrac{4}{x^2} = 0\) at \(x = 2\)
\(f(1) = 5\)
\(f(2) = 4\)
\(f(4) = 5\)
Answer: max 5 at the endpoints; min 4 at \(x = 2\)
The only critical point in the interval is \(x = 0\), where \(f'\) is undefined.
\(f(-1) = 1\)
\(f(0) = 0\)
\(f(8) = 4\)
Answer: max 4 at \(x = 8\); min 0 at \(x = 0\)
\(T'(x) = 6 - 2x = 0\) at \(x = 3\)
\(T(0) = 20\)
\(T(3) = 20 + 18 - 9 = 29\)
\(T(5) = 20 + 30 - 25 = 25\)
Answer: highest 29°C; lowest 20°C
\(f'(x) = 3x^2 - 3 = 0\) at \(x = \pm 1\)
\(f(-2) = -2\)
\(f(-1) = 2\)
\(f(1) = -2\)
\(f(2) = 5\)
Answer: 5 at \(x = 2\)
The smallest candidate value is 4.
Answer: 4
A product with an exponential, a logarithm, or an absolute value.
\(f'(x) = e^{-x}(2x - x^2) = x e^{-x}(2 - x)\)
Candidates in the interval: \(x = 0\) and \(x = 2\), plus the endpoint \(x = 4\).
\(f(0) = 0\)
\(f(2) = \dfrac{4}{e^2}\)
\(f(4) = \dfrac{16}{e^4}\)
\(\dfrac{4}{e^2} > \dfrac{16}{e^4}\) because \(e^2 > 4\).
Answer: max \(\dfrac{4}{e^2}\) at \(x = 2\); min 0 at \(x = 0\)
\(f'(x) = \dfrac{1 - \ln x}{x^2}\)
\(\ln x = 1\) at \(x = e\), which is in the interval.
\(f(1) = 0\)
\(f(e) = \dfrac{1}{e}\)
\(f(e^2) = \dfrac{2}{e^2}\)
\(\dfrac{1}{e} > \dfrac{2}{e^2}\) because \(e > 2\).
Answer: max \(\dfrac{1}{e}\) at \(x = e\); min 0 at \(x = 1\)
On \([-1, 2]\), \(f(x) = 4 - x^2\) and \(f'(x) = -2x\), so \(x = 0\) is a candidate.
On \([2, 3]\), \(f(x) = x^2 - 4\) and \(f'\) is not 0.
The expression inside the absolute value is 0 at \(x = 2\), so the derivative fails there.
\(f(-1) = 3\)
\(f(0) = 4\)
\(f(2) = 0\)
\(f(3) = 5\)
Answer: max 5 at \(x = 3\); min 0 at \(x = 2\)
\(f\) is continuous on \([-1, 2]\).
The only critical point is \(x = 0\).
The candidate values are \(f(-1) = 1\), \(f(0) = 0\), and \(f(2) = 4\).
The absolute maximum value is 4, at \(x = 2\).
Answer: absolute max 4 at \(x = 2\)
The candidate values are 1, 5, 1, and 5.
The smallest is 1.
Answer: A. \(1\)
\(f(0) = 0\) and \(f(4) = 16/e^4\)
\(f(2) = \dfrac{4}{e^2}\)
\(\dfrac{4}{e^2}\) is larger than both endpoint values because \(e^2 > 4\).
The absolute maximum is at the critical point \(x = 2\).
Answer: the critical value \(\dfrac{4}{e^2}\) is larger than both endpoint values
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