f'(x)=x(e^x-2)

Let f' (x) > 0, < 0, =0 respectively.

When x∈(-∞, 0)∩(LN2, +∞), f' (x) > 0.

F'(x)=0 when x=0 or x=ln2.

F' (x) < 0 when x∈(0, ln2).

To sum up, f(x) increases at (-∞, 0), decreases at (0, ln2) and increases at (ln2,+∞).

There is a maximum value of-1 when x=0, and a minimum value of ln4-(ln2) when x=ln2. -2.

(2)

Let f(x)=0 and get k = (x- 1) e x/x? Then let g (x) = (x- 1) e x/x? .

Therefore, the problem is transformed into proving that there is only one intersection between g(x) and y=k images.

g'(x)=2e^x(x？ -x+ 1)/x？

∵e^x,x？ For x∈R, x+ 1 is greater than 0.

When x > 0, g' (x) > 0, and when x < 0, g' (x) < 0.

∴g(x) decreases at (-∞, 0) and increases at (0,+∞).

Let g (x) < 0 and g (x) > 0 respectively, and the solution is:

G (x) < 0 when x < 0.

When 0 < x < 1, g (x) < 0.

G (x) > 0 when1< x.

When ∴x∈[ 1, +∞), g(x)≥0.

And ∵g(x) monotonically increases at (0, +∞).

The intersection of ∴g(x) and y=k can only be 1, and only x∈[ 1, +∞) has a unique solution.

To sum up, it proves that.