The known function f x xlnx 20

Answer: f(x) = xlnx

df/dx = lnx + 1

Ream: df dx > 0, get: lnx + 1 > 0, lnx > 1, x > 1 e

Answer: The monotonous upward range is (1 e, +.)

The monotonic descent interval is (0, 1 e).

g(x) = f(x) +f(k-x)

xlnx + k-x)ln(k-x)

dg/dx = lnx + 1 - ln(k-x) -1ln[x/(k-x)]

Let dg dx = 0, then ln[x (k-x)] = 0x (k-x) = 1, x = k - x, x = k 2d 2g dx 2 = 1 x + 1 (k-x) when x = k 2

d 2g dx 2 = 2 k + 1 (k-k 2) = 4 k f(x) has a minimum value at k > 0.

fMin = (K2)LN(K2) +K-K2)Ln(K-K2).

kln(k/2)

Solution (1) f'(x)=1+lnx, let f'(x) >0 then x>1'(x)<0, then x(0,1 e).function f(x).

The monotonically increasing interval is (1 e,+ the monotonically decreasing interval is (0,1 e)2)g(x)=f(x)+f(k-x)=xlnx+(k-x)ln(k-x), and the domain is defined as (0,k),k>0, otherwise the function g(x) is meaningless. g'(x)=lnx/(k-x),g'(x) >0, then x (k-x) > 1, k 2g'(x)<0, then 0g(x) in x=k2 is the minimum value obtained for kln(k2).

1. The domain of the function is defined as: [0, positive infinity].

Derivative: y'=log(x) +1=0

x=1 e let x1<1 e, when, y'is less than 0, so the function has a decreasing interval [0,1 e] and an increasing interval [1 e,infinity].

2、g(x) =x*log(x) -k - x)log(k -x )g'(x)= log(x - k) +log(x) +2

f(x)=x-xlnx

Deriving step by step.

First to the previous x

f'(x)=1-(xlnx)'

Then transport the reed to the back of the XLNX to find the derivation (the front car quietly old guide after the guide + after the closed guide before the guide is not conducted) f'(x)=1-(xlnx)'

1-(lnx+1)

lnx

y=x^lnx

Logarithmic derivative:

Take the logarithm of both sides at the same time to get :

lny=(lnx)^2

Derivation: y'/y=2lnx/x

y'=2x^(-1)(lnx)x^lnx

y'=2(lnx)x^(lnx-1)

Solution: (1) Find the derivative of the function, and use the geometric meaning of the derivative to find the equation of the tangent l l at the point p(1,f(1)) of the image of the function y=f(x);

2) Solve the logarithmic inequality to find it;

3) From the meaning of the problem, the equivalent of k f(x) x 1]=[x+xlnx x 1] is constant for any x 1, so that g(x)=[x+xlnx x 1], and the minimum value of the function can be obtained by using the derivative

1), when a=1 f(x)=x+xlnx f (x)=2+lnx, f (1)=2, f(1)=1, the tangent equation is y-1=2(x-1), i.e. y=2x-1....(2 points).

2) f(x)=ax+xlnx, and the domain of the function is (0,+ f(x) 0 t'ncore a+lnx 0, x (0,e-a)....(4 points).

3) When a=1, the line y=k(x-1) is constant below the image of the function y=f(x) for x(1,+), and the problem is equivalent to k

f(x)x 1=[x+xlnx x 1]for any x 1 constant true....(5 points).

Let g(x)=[x+xlnx x 1], g (x)=[x 2 lnx

x 1)2, so that h(x)=x-2-lnx, so h(x) is an increasing function on (1,+, since h(3)=1-ln3 0, h(4)=2-ln4 0

So there is x0 (3,4) such that h(x0)=x0-2-lnx0=0

then x (1, x0) and h(x) 0;x (x0, +, h(x) 0, i.e. x (1, x0), g'（x）＜0；x (x0, +, g.)'（x）＞0

It is known that g(x) is decreasing at (1,x0), (x0,+ increasing....(10 points).

and g(x0) g(3)=

3 2(LN3+1) G(4)=2+2LN4, so KMAX=3 ....(12 points).

Comments: Test points for this question: The application of derivatives in the maximum and minimum value problems

Test Center Comments: This question mainly examines the use of derivatives to study the tangent equation of the function, monotonicity, maximum value and other properties, and tests the student's arithmetic ability.

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Wonderful. copyright © 2021 yulucn. -17 q. s.-webmaster@ , known function f(x)=ax+xlnx

1) When a=1, the tangent equation of the image of the function f(x) at the point p(1,f(1));

2) When a 0, the solution of the inequality f(x) 0;

3) When a=1, find the maximum value of the integer k below the image of the function y=f(x) for x (1,+ straight line operation y=k(x-1).

The original letter imitates the number of lines f'(x)=f(prepared with notice x).

f(x)=xlnx

f(x)=f‘(x)=lnx+1

f'(x)= 1/x

1. Tangent equation x=e point y=f(e)=elne=e

Slope k=f'(x)=lne+e/e=2 y=f(x)=2（x-e)+e=2x-e

2. f(x)=f(x) a=xlnx a derivative (lnx+1) a a>0 So the reciprocal is an increasing function.

x belongs to [a,2a].

lna+1）/a (ln2a+1)/a

lna+1) a >0 a>1 e derivative greater than 0 f(x) is an increase, and the maximum value is 2ln(2a).

ln2a+1) a<0 0< a<1 (2e) the derivative is less than 0 f(x) is minus, and the maximum is ln(a).

x(0,+ assumes xlnx>x e x-2 e --x(lnx-e (-x))>2 e

Let g(x)= x(lnx-e (-x)) derive lnx-e (-x)+1+e (-x)=lnx+1

When lnx+1>0 i.e. x>1 e g(x) is incremented.

When lnx+1<0 is 00 i.e. -2 e

XLNX>X E X-2 E was established.

1. Derivative, get f'(x)=(xlnx)'=lnx 1, so the slope of the tangent k=f'(e) = 2, and the tangent coordinates are (e, e).

2、f'(x) = (1 a) (lnx 1), since a>0, so f'(x)>0 is constant over the interval [a,2a], i.e., f(x) increases monotonically over the interval so that the maximum value is f(2a).

3. It should be a post-variant constructor, using the derivative to determine the monotonicity of the new function, and then proving its minimum value of 0. . . The idea should be like this, the structure is easy to deal with, hehe. It should belong to the finale question type of the senior three comprehensive paper.

1.The derivative can be used to find the monotonic interval of a function.

f'(x)=-1/(xlnx)^2*(lnx+1)=-lnx+1)/(xlnx)^2

f'(x)=0

1+lnx=0

lnx=-1

x = 1 e When x < 1 e, f'(x) >0, the function is an increment.

When x>1 e, f'(x) <0, the function is a subtraction.

2.You can take the logarithm on both sides at the same time.

Both sides take the logarithm ln2 x> alnx to x (0,1) is true, i.e. 1 xlnx-eln2

It proves that lnx 2(x-1) (x+1) is true at x 1. Let p(x)=lnx-2(x-1) (x+1).

Derivative p(x)'=x-1) 2 x(x+1) 2>0 at x 1. So monotonically incremental. So p(x)>p(1)=0So LNX-2(X-1) (X+1) >