Mathematical derivative

The function f (x) = ax+x2-xlna (a >; 0，a≠ 1)。

(1) When a >: at 1, it is proved that the function f(x) increases monotonically at (0, +∞);

(2) If the function y = | f (x)-t |- 1 has three zeros, find the value of t;

(3) If there are x 1, X2 ∈ [- 1, 1], let | f (x1)-f (x2) | ≥ e-1,and try to find the value range of A.

(1) see analysis (2)t = 2(3)∩[E, +∞)

Guide to examination: This question examines the comprehensive properties of function and derivative, and the function model is not complicated. (2) The two questions are very routine. We should use derivatives to test the monotonicity of functions and equations. (3) We should convert "if there is x 1, X2 ∈ [- 1" into | f (x1)-f (x2) | ≥ e-1".

Standard solution: (1) Proof: f ′ (x) = axlna+2x-lna = 2x+(ax-1) lna. (2 points)

Because of a> 1, when x∈(0, +∞), LNA >；; 0，ax- 1 & gt； 0, so f' (x) > 0.

Therefore, the function f(x) monotonically increases on (0, +∞). (4 points)

(2) solution: when a>0, a≠ 1, because f ′ (0) = 0 and f ′ (x) monotonically increases on r, f ′ (x) = 0 has a unique solution X = 0. (6 points), so the changes of x, F'(X) and f(x) are shown in the following table.

(-∞,0)

(0,+∞)

f′(x)

f(x)

minimum value

The function y = | f (x)-t |- 1 has three zeros, so the equation f (x) = t 1 has three roots, t+ 1 >: T- 1, so t-1. (10).

(3) solution: because there is x 1, X2 ∈ [- 1, 1], making | f (x1)-f (x2) | ≥ e-1,so when x.

According to (2), f(x) decreases at [- 1, 0] and increases at [0, 1], so when x ∈ [- 1, 1], f (x) min.

And f (1)-f (-1) = (a+1-lna)-= a-2 lna,

Remember that g (t) = t-2lnt (t >; 0), because g' (t) = 1+-= ≥ 0 (if and only if t = 1 takes the equal sign),

So g (t) = t-2lnt monotonically increases on t∈(0, +∞), while g (1) = 0.

So when t & gtAt 1, g (t) >: 0; When 0