Given the function f (x) = 3ax + 1-2a, there exists X1 on [- 1,1], such that f (x1) = 0 (x1 is not equal to plus or minus one), then the value range of a is () A.（-1,1/5） B.（1/5,+∞） C.（-∞,-1）∪（1/5,+∞） D.（-∞,-1）

Given the function f (x) = 3ax + 1-2a, there exists X1 on [- 1,1], such that f (x1) = 0 (x1 is not equal to plus or minus one), then the value range of a is () A.（-1,1/5） B.（1/5,+∞） C.（-∞,-1）∪（1/5,+∞） D.（-∞,-1）

f(-1)f(1)1/5,a

Understanding of "roots of equations and zeros of functions" in teaching parameters
It is mentioned in the teaching reference: "after giving the concept of zero point of function, let the students know that 'root of equation' and 'zero point of function' are closely related, but they can not be confused." how to understand this? Is the word 'root of equation' not limited to the root of one variable equation?
Thank you for your answer, but you don't seem to know what the zero point of a function is. Please check it in the compulsory 1 textbook
Thank you for your reply on the second floor. You make a lot of sense. Let me see
Thank you for your reply on the third floor

I think what the author means is that sometimes the zero point of the function may not exist, but the root of the equation can exist, for example
The equation x ^ 2 + 1 = 0 has roots, but in the complex field
But the function x ^ 2 + 1 = y has no zero
Personal opinion, hope to help the landlord
Respect the great teacher, I hope I can be a teacher in the future

Roots of equations and zeros of functions
If the function f (x) is monotone in its domain of definition, then f (x) has at most one zero point. Given that f (x) = a ^ x + (X-2) / (x + 1) (a > 1), we try to prove that the equation f (x) = 0 has no negative root
Our teacher said that the first sentence is not right, directly use the following conditions, thank you_ ∩)o...
How to prove

If the function f (x) is monotone in its domain, then f (x) has at most one zero point. This is true. Discontinuity is also at most one zero point. However, only when f (x) is continuous in the domain, can we judge the zero point accurately. First, we prove that f (x) is monotone, f '(x) = a ^ xlna + 3 / (x + 1) ^ 2 > 0, and the break point x = - 1, so the piecewise increment x → - 1 (...)

On the roots of equations and the zeros of functions
1. It is known that the real numbers a, B and C are three numbers in the domain of y = f (x), and satisfy a
Can you omit a little bit?
We didn't learn dichotomy! Is there another way?
No,
Just answer the first three questions!

Have you learned derivative? Others are truncated string method, Newton downhill method, iterative method
2. I have sent out the answers to three questions