2. 若兩個函數f和g在開區間(a,b)下凹(Concave Downward)，則f+g在開區間(a,b)仍然為下凹。
3. 若����''(5) = 0，則函數f在x5處必為反曲點(Inflection Point)。
1. 請問相對極大值(Relative Maxima)是否有可能比相對極小值(Relative Minima)小？試畫
2. 請問臨界點(Critical Point)是否一定為相對極值(Relative Extrema)？試舉例說明之。
3. 請找出函數f(x) =1/x3(x的三次方)的相對極值和反曲點
- ¥¿¶i ¦Lv 79 年前最佳解答
proof:Let f : R→R and g : R→R be increasing functions. We will show that f + g is also an increasing function. Let a,b∈R, where a < b. Since f is increasing and a < b, we have that f(a) < f(b). Similarly, since g is increasing and a < b, g(a) < g(b). Thus,(f + g)(a) = f(a) + g(a) < f(a) + g(b) (since g(a) < g(b))
< f(b) + g(b) = (f + g)(b) (since f(a) < f(b))
Hence, (f + g)(a) < (f + g)(b) whenever a < b, which shows that f + g is an increasing function.
Assume that f and g are both concave downward at x = a, then we know that f "(a) and g"(a) are both negative. The second derivative of f is f "(x) + g"(x).Since we know that at x = a ,f "(a) and g"(a) are negative, then f "(a) + g"(a), must also be negative. Hence f + g must be concave downward at x = a.
note that f(x)=(x-5)^4, f '(x)=4(x-5)³,f "(x)=20(x-5)².����''(5) = 0,
(5,f(5)) is not an inflection point! In fact f(5)=0 is a mininmum value!
1.A local minimum value may be greater than a local maximum value.
See the figure 1.
2.We take a counterexample! Let f(x)=x³, clearly x=0 be a critical point, f(0)=0 is not a Relative Extrema. in actually, (0,0) is an inflection point.