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(20 14? The function f(x)=lnx-mx(m∈R) is known. (1) If curve y= ...
Solution: Solution: (1) Because point P( 1,-1) is on the curve y=f(x),

So -m=- 1, the solution is m= 1.

Because f '(x)= 1

1

x

- 1=0,

So the slope of the tangent is 0,

So the tangent equation is y=- 1.

(2) Because f '(x)= 1

1

x

-m=

1-mx

x

.

① when m≤0, x∈( 1, e) and f ′ (x) > 0,

Therefore, the function f (x) monotonically increases at (1, e),

Then f (x)max=f (e)= 1-me.

② When?

1

m

≥e, that is, 0 < m ≤

1

e

X∈( 1,e),f′(x)> 0,

Therefore, the function f (x) monotonically increases at (1, e),

Then f (x)max=f (e)= 1-me.

③ When 1

1

m

< e, i.e.

1

e

< m < 1,

The function f (x) is in (1,

1

m

) monotonically increasing, in (

1

m

E) monotonically decreasing,

Then f (x)max=f (

1

m

)=-lnm- 1。

④ When?

1

m

≤ 1, that is, when m≥ 1, x∈( 1, e), f ′ (x) < 0,

The function f (x) monotonically decreases at (1, e),

Then f (x) max = f (1) =-m.

To sum up, ① when m≤

1

e

f(x)max = 1-me;

② When?

1

e

< m < 1,f(x)max =-lnm- 1;

③ When m≥ 1, f (x) max =-m. 。

(3) let x 1>x2>0 > X2 > 0.

Because f (x 1)=f (x2)=0,

So lnx 1-mx 1=0, lnx2-mx2=0,

Lnx 1+lnx2=m(x 1+x2),lnx 1-lnx2=m(x 1-x2)。

To prove that x 1x2 > E2,

That is, lnx 1+lnx2 > 2, that is, m (x 1+x2) > 2.

Because m=

lnx 1-lnx2

x 1-x2

So this proves that

lnx 1-lnx2

x 1-x2

>

2

x 1+x2

Namely ln

x 1

x2

>

2(x 1-x2)

x 1+x2

.

manufacture

x 1

x2

=t, then t > 1, then lnt > >

2(t- 1)

t+ 1

.

Order? (t)=lnt-

2(t- 1)

t+ 1

(t> 1),

then what ′(t)=

1

t

-

four

(t+ 1)2

=

(t- 1)2

t(t+ 1)2

>0.

What about the function? (t) Incremental function on (1, +∞),

So what? (t)>? (1)=0, i.e. lnt > >

2(t- 1)

t+ 1

Established.

So the original inequality holds.