吾該有沒人識做呢幾條問題啊？

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圖片參考：http://upload.wikimedia.org/wikipedia/commons/0/02/Quasiconvex_function.png


This graph is a quasiconvex function which is not convex.

Now, you reflect the graph with the x-axis, then the resluting graph is a quasiconcave function that is not concave.

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Proof
Denote the function by f , and the (convex) set on which it is defined by S. Let a be a real number and let x and y be points in the upper level set Pa: x ∈ Pa and y ∈ Pa. We need to show that Pa is convex. That is, we need to show that for every λ ∈ [0,1] we have (1 − λ)x + λy ∈ Pa.
First note that the set S on which f is defined is convex, so we have (1 − λ)x + λy ∈ S, and thus f is defined at the point (1 − λ)x + λy.
Now, the concavity of f implies that

f ((1−λ)x + λy) ≥ (1−λ) f (x) + λ f (y).Further, the fact that x ∈ Pa means that f (x) ≥ a, and the fact that y ∈ Pa means that f (y) ≥ a, so that

(1−λ) f (x) + λ f (y) ≥ (1−λ)a + λa = a.Combining the last two inequalities, we have

f ((1−λ)x + λy) ≥ a,so that (1−λ)x + λy ∈ Pa. Thus every upper level set is convex and hence f is quasiconcave
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圖片參考：http://in.geocities.com/myisland8132/yahooknowledge/7007102000410.bmp

The gradient is (-2x,-2y)
So the gradient at (1,1) is (-2,-2)
you just need to draw the vector (-2,-2) at (1,1)