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CA asked in 科學數學 · 10 years ago



是非題 (答案若為是,請說明之;答案若為否,則請舉反例說明)

1. 若兩個函數f和g在開區間(a,b)遞增(Increasing),則f+g在開區間(a,b)仍然為遞增。

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

1 Answer

  • 10 years ago
    Favorite Answer



    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.




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