Elementary Calculus: Direct Test

DIRECT TEST

Suppose c is the only interior critical point of f, and u, v are points in I with u < c < v.

(i) If f(c) > f(u) and f(c) > f(v)> then g has a maximum at c and nowhere else.

(ii) If f(c) < f(u) and f(c) < f(v), then f has a minimum at c and nowhere else.

(iii) Otherwise, f has neither a maximum nor a minimum at c.

The three cases in the Direct Test are shown in Figure 3.5.6. The advantage of the Direct Test is that one can determine whether f has a maximum or minimum at c by computing only the three values f(u),f(v), and f(c) instead of computing all values of f(x).

PROOF OF THE DIRECT TEST

We must prove that if two points of I are on the same side of c, their values are on the same side of f(c). Suppose, for instance, that u₁ < u₂ < c (Figure 3.5.7). On the closed interval [u₁, c] the only critical points are the endpoints. Thus when we restrict f to this interval, it has a maximum at one endpoint and a minimum at the other. If the maximum is at c, then f(u₁) and f(u₂) are both less than f(c); if the minimum is at c, then f(u₁) and f(u₂) are both greater than f(c). A similar proof works when c < v₁ < v₂.

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Direct Test DIRECT TEST Suppose c is the only interior critical point of f, and u, v are points in I with u < c < v. (i) If f(c) > f(u) and f(c) > f(v)> then g has a maximum at c and nowhere else. (ii) If f(c) < f(u) and f(c) < f(v), then f has a minimum at c and nowhere else. (iii) Otherwise, f has neither a maximum nor a minimum at c. The three cases in the Direct Test are shown in Figure 3.5.6. The advantage of the Direct Test is that one can determine whether f has a maximum or minimum at c by computing only the three values f(u),f(v), and f(c) instead of computing all values of f(x). Figure 3.5.6 PROOF OF THE DIRECT TEST We must prove that if two points of I are on the same side of c, their values are on the same side of f(c). Suppose, for instance, that u₁ < u₂ < c (Figure 3.5.7). On the closed interval [u₁, c] the only critical points are the endpoints. Thus when we restrict f to this interval, it has a maximum at one endpoint and a minimum at the other. If the maximum is at c, then f(u₁) and f(u₂) are both less than f(c); if the minimum is at c, then f(u₁) and f(u₂) are both greater than f(c). A similar proof works when c < v₁ < v₂. Figure 3.5.7
Home Continuous Functions Maxima and Minima Interior Points of Intervals Direct Test