Honey, you should listen carefully. Here is a brief description for everyone. First, the function image passes through the point (0, 1), indicating that when x is equal to 0, f(x)= 1. Ok, substituting in the expression, we can see that both variables A and B become 0, which means that C is 1. Then, next, we take the derivative of the function, and the derivative of a point is the slope of the tangent of that point. Then the derivative result is f' (x) = 4ax 3+2bx. Next, we need to find two equations to solve the values of A and B. According to the topic, there is a hidden slope in the tangent equation, which is an equation and a hidden depth, which is the tangent at x= 1, that is to say, when x= 1, the value of y is also certain (the tangent and the original function pass through the same point), and this value is-/. So the first equation is 4a+2b= 1, the left: x= 1, and the right: the expression of slope when the slope is 1. The second equation is a+b+ 1=- 1 (simplified as a+b=-2), where. The specific method is that the second equation is multiplied by 2 to become the first equation, then a=5/2, and then b=-9/2). So f (x) = 5/2 * (x 4)-9/2 * (x 2)+1 Bring it back for inspection. Ok, that's all for the time being. I won't ask again.