The following functions satisfy the condition of Lagrange mean value theorem on [1, e] is (a lnlnx B LNX C 1 / LNX D ln (2-x)

The following functions satisfy the condition of Lagrange mean value theorem on [1, e] is (a lnlnx B LNX C 1 / LNX D ln (2-x)

Choose B
The condition of Lagrange mean value theorem is:
(1) Continuous (2) on [1, e] is differentiable on (1, e)
Look at the answers one by one
A. There is no definition on x = 1, so it is discontinuous and wrong in [1, e]
C. Same as a
D. Since E > 2, there is no definition on X belonging to (2, e), which is wrong
Only B can qualify

Lagrange mean value theorem when x > 0, ln (1 + 1 / x) > 1 / (1 + x)

Let f (x) = LNX
There is y ∈ (x, x + 1) such that
f'(y)=[f(x+1)-f(x)]/(x+1-x)
=ln(x+1)-lnx
=ln(1+1/x)
∵0

When x > 1, it is proved that ex > ex

In order to prove that when x > 1, ex > ex, it only needs to prove ex ex > 0
Let f (x) = ex ex, then f (1) = 0
Because f '(x) = ex-e,
Therefore, when x > 1, f '(x) > 0,
Thus, f (x) > F (1) = 0,
That is, when x > 1, ex ex > 0

It is proved that when x > 1, the x power of E is greater than Xe

prove:
Let f (x) = e ^ x-xe
be
f'(x)=e^x-e>0 (x>1)
So f (x) is strictly increased
Therefore, f (x) ≥ f (0) = 1 > 0
thus
e^x>ex

E ^ x > 1 + X, X ≠ 0 prove inequality

prove:
Constructor f (x) = e ^ x-1-x
f(0)=e^0-1-0=0
f'(x)=e^x-1
When x > 0 and f '(x) > 0, f (x) increases
When x

When x > 0, the inequality ex > 1 + X + 1 is proved 2x2 established

Proof: Let f (x) = ex − 1 − x − 1
2x2，
Then f '(x) = ex-1-x,
Let g (x) = f '(x), then G' (x) = EX-1,
∵ x ＞ 0, ∵ EX-1 ＞ 0, i.e. G '(x) ＞ 0,
‡ g (x) is an increasing function on [0, + ∞),
Since x > 0, then G (x) > G (0) = e0-1 = 0, that is, f '(x) > 0,
‡ f (x) is an increasing function on [0, + ∞),
From x > 0, f (x) > F (0) = E0 − 1 − 0 − 1
two × 02＝0，
Ex - (1 + X + 1)
2x2）＞0，
∴ex＞1+x+1
2x2, certified