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Critical point (in one variable): A point on the graph y = f(x) at which f is differentiable and f'(x) = 0. The term is also used for the number c such that f'(c) = 0. The corresponding value f(c) is a critical value. A critical point c can be classified depending upon the behavior of fin the neighborhood of c, as one of following:
1. a local minimum, if f'(x) > 0 to the left of c and f'(x)< 0 to the right of c.
2. a local maximum, if f'(x) < 0 to the left of c and f'(x)> 0 to the right of c.
3. neither local maximum nor local minimum:
a) if f'(x) has the same sign to the left and to the right of c, in which case c is a horizontal point of inflection.
b) if there is an interval at every point of which f'(x) = 0 and c is an endpoint or interior point of this interval.
(This is called the first derivative test).
The second derivative test (is also a test for maximum and minimum values. It is a consequence of the concavity test):
Suppose f is continuous near c.
1. If f'(c) = 0 and f''(c) > 0, then f has a local minimum at c.
2. If f'(c) = 0 and f''(c) < 0, then f has a local maximum at c.
Example: f(x) = x^3 - 12x + 1
(a) Find the intervals on which f is increasing or decreasing.
(b) Find the local maximum and minimum values of f.
(c) Find the intervals of concavity and the inflection points.
Solution:
(a) f(x) = x^3 - 12x + 1
f'(x) = 3x^2 - 12
f'(x) = 3(x +2)(x - 2)
Interval: x < -2; -2 x < 2; x > 2
x + 2: - ; +; +
x - 2: - ; - ; +
f'(x): + ; - ; +
f: increasing on (-∞, -2); decreasing on (-2, 2); increasing on (2, ∞) So f is increasing on (-∞, -2) and (2, ∞) and f is decreasing on (-2, 2).
(b) f changes from increasing to decreasing at x = -2 and from decreasing to increasing at x = 2. Thus f(-2) = 17 is a local maximum value and f(2) = -15 is a local minimum value.
(c) f''(x) = 6x
f''(x) > 0 ↔ x > 0 and f''(x) < 0 ↔ x < 0. Thus f is concave upward on (0, ∞) and concave downward on (-∞, 0). There is an inflection point where the concavity changes, at (0, f(0)) = (0, 1).
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