f(x)=xe^ax- 1/x,x≠0
f(x)=x(e^ax- 1/x? )x-& gt; 0+,f(x)-& gt; -∞; x-& gt; 0-,f(x)-& gt; +∞, x=0 is a breakpoint, but not a zero point.
F(x) has two zeros, which is equivalent to g (x) = e ax- 1/x? There are two zeros.
For example, y = (e a) x and y= 1/x? Intersection point, as shown in the figure below:
A = 0,e 0 = 1,y = (e A) x = 1 x = 1,y= 1/x? There is an intersection on each side;
e^a>; 1=e^0,a>; 0 and one
As shown above, there must be an intersection point on the right side of the Y axis;
On the left side of the y axis, x
When tangent, the slopes of two curves are equal at the tangent point. Tangent points are also intersections.
Let the intersection (x0, y0), x0.
y 1=e^ax,y 1'=ae^x,>; 0y0=e^ax0
y2= 1/x? ,y2'=-2/x? & gt0,
ae^x0=-2/x0? ;
y0=e^ax0
y0= 1/x0?
ay0=-2/x0? ;
a=(-2/x0? )/( 1/x0? )=-2x0,x0=-a/2
y0= 1/(a? /4)=4/a?
ay0=-2/(-a? /8)= 16/a? ,y0= 16/a^4
4/a? = 16/a^4
1=4/a?
Answer? =4,a=2
∴-2≤a≤2