YLL討論網

1.  Since f(0) = 5 > 0, f(9) < 0, there must be at least one root between 0 and 9.
2. When x approaches negative infinity, f(x) will be negative. This implies that there exists one negative root.
3. When x approaches positive infinity, f(x) will be positive. This means that there is one root greater than 9.


So three real roots. (Sorry in English. If Chinese characters were used, the output was just not right.)

1.  Since f(0) = 5 > 0, f(9) <0> Three real roots. (Sorry in English. If Chinese characters were used, the output was just not right.)

[quote="cho"]a,b屬於實數
f(x)=x^3+ax^2+bx+5
若f(9)<0> 0, f(9) < 0：所以在 0<x<9 之間一定有一根
2. x 趨近負無限大時, f(x) <0： 所以在x<0>0： 所以在x>9, 必定也有一根

==> 共有三個實根.

a,b屬於實數
f(x)=x^3+ax^2+bx+5
若f(9)<0
則方程式f(x)=0 有幾個實根？

答案為3
但不知道如何解
煩請解答 謝謝