High and high derivatives questions, please ask the experts to answer

Let f(x)=2x 3-3(a-1) 2*x+1 (a>=1).

1) Find the single increment interval of f(x).

Analysis: f(x)=2x 3-3(a-1) 2*x+1 (a>=1).

Let f'(x)=6x 2-3(a-1) 2=0==>x1=- 2 2*(a-1),x2= 2 2*(a-1).

f'(x) is the parabola with an opening upward, which changes from positive to negative when passing x1, and f(x) takes the maximum value at x1; When x2 is crossed, it changes from negative to positive, and f(x) is taken as the minimum value at x2.

When a>1.

When x (-x1), f'(x) >0 and f(x) increase monotonically; x [x1, x2), f'(x)<0, f(x) monotonically decreased; x [x2,+,f'(x)>0, f(x) monotonically increases;

When a>1.

x (-, f'(x)>0, f(x) monotonically increases;

2) Discuss the extreme value of f(x).

From (1) to know when a>1.

At x1, f(x) takes the maximum value f(x1); At x2, f(x) takes the minimum f(x2).

When a=1, there is no extremum.

1, f(x) derivation yields 6x 2-3(a-1)2. Obviously, the derivative function is a parabola, and it is not difficult to get 2 increasing intervals and 1 decreasing interval.

2. The extreme value occurs when the function transitions from the increasing interval to the decreasing interval (maximum).

Or transition from the subtracting interval to the increasing interval (minimum), and calculate the specific number by yourself (all expressions of a).

Derivative = 6x 2-3(a-1) 2 6x 2-3(a-1) 2=0 to get x = plus or minus half of the change sign 2*(a-1).

So the single increment is (negative infinity, minus half change sign 2*(a-1)), half change sign 2*(a-1), positive infinity).

The extremum can bring in the two values obtained by the solution of x.

for the number of tours in Changzhou) y'=0 y'=nx^(n-1) y'=a^xlna y=e^x y'=e^x

y'=logae/x y=lnx y'Book abrasive pin = 1 x y'=cosx y'=-sinx

y'=1/cos^2x y'=-1/sin^2x

2 algorithms.

The law of addition (subtraction): [f(x)+g(x)].'f(x)'+g(x)'Call.

Multiplication: [f(x)*g(x)].'f(x)'*g(x)+g(x)'*f(x)

Division rule: [f(x) g(x)].'f(x)'*g(x)-g(x)'*f(x)]/g(x)^2

Table of derivatives of basic elementary functions.

y'=0 '=μ1) y'=a^x lna y=e^x y'=e^x

x y'=loga,e/x y=lnx y'=1/x y'=cosx

y'=-sinx y'=(secx)^2=1/(cosx)^2

y'=-cscx)^2=-1/(sinx)^2 sinx y'=1/√(1-x^2)

cosx y'=-1/√(1-x^2) tanx y'=1/(1+x^2)

cotx y'=-1/(1+x^2) x y'=ch x

x y'=sh x y'=1/(chx)^2

shx y'=1/√(1+x^2)

chx y'=1/√(x^2-1) th y'=1/(1-x^2)

1.is an even function.

f(x) is the increment function.

Solution: The increasing interval of the function y= is (-2).

Because it is less than 1, the peripheral function is a subtraction function, and now you only need to find the subtraction interval of the inner function, so the increasing interval is negative infinity to negative 2

The function y=log (is the logarithm of the base less than 1 and is a subtraction function. And the function u=(x+2) 2 decrements at x<-1. The law of monotonicity of composite functions:

Same increase and different subtraction. So the increment interval of the function is x<-1.If we remove the exponent and get y=2log(, we change the definition domain, so that the decrease becomes a decrease without an increasing interval.

First, use the formula for changing the bottom, the original formula = (2 2|, to derive y'=(2/ 2|(x>=-2), y'=-(2/ 2|(x<-2).When y'>=0 the function increases monotonically, so the increment interval is (-infinity, -2].

Because < 1, the outer function is a subtraction function, then it depends on the internal test, when the function is increasing as a whole, the inner side is a subtraction function, and its subtraction interval is (- 2).

Haha, no one has robbed me of this question, let's see who dares to rob me. As soon as the simple question of the province came out, it was robbed of the head by some small pot friends. Depressed! Hehe!

If there is a tangent line y=kx+m-k, then the intersection point (x,y) must satisfy f(x) and the ordinate of the line to be equal, and the slope should also be equal.

x -3x = kx + m-k, and 3x -3 = k means that when m takes some value, x has three solutions.

Substituting to get: 2x -3x +m+3=0 to have 3 roots to the left derivation: 6x -6x

So at the minimum, x=1, left=m+2

x = 0 at maximum, left = m+3

The equation has three roots, i.e., the maximum value of 0 and the minimum value <0

m+2<0∴-3

f(x)=ln(x+1)-ax/(x+a)

a>1)

f(x)=ln(x+a) lnx,a=1, find the monotonic interval of f(x).

Solution: f(x)=ln(x+1) lnx when a=1; Define domains: x>0 and x≠1;

Since f(x)=[lnx) (x+1)-(1 x)ln(x+1)] ln x=[xlnx-(x+1)ln(x+1)] x(x+1)ln x]<0

In its defined domain x>0 is constant, so that f(x) is monotonically reduced in (0,1) (1,+).

x→0limf(x)=x→0lim[ln(x+1)/lnx]=0；

x→1⁻limf(x)=x→1⁻lim[ln(x+1)/lnx]=-

x→1⁺limf(x)=x→1⁺lim[ln(x+1)/lnx]=+

x→+∞limf(x)=x→+∞lim[ln(1/x)/lnx]=x→+∞lim[x/(x+1)]=1.

（1）f(x)=x²+lnx-ax（a∈r）f'(x)=2x+1/x-a

f(x) increments the function on (0,1), indicating that on (0,1), f'(x) 02x+1 x 2 2, and the equal sign is obtained at x = 2 2 2 so 2x+1 x-a 0

a 2x+1 x minimum.

a≤√2/2

The value range of a is (- 2 2].

2）x∈[0，ln3]

t=e^xt∈[1，3]

Because a 2 2 < 1

So |e^x-a|=e^x-a

g(x)=e^2x+e^x-a

g'(x)=2e^2x+e^x>0

g(x) is an increment function on the defined domain.

The minimum value of g(x)min=g(0)=1+1-a=2-a function g(x) is 2-a

1. When a=1, f(x)=x -x-ln(x-1)f'(x)=2x-1-1 (x-1).

f'(x)=0, there is x=

So the maximum value of f(x) is f(

2、f‘（x）=2x-a-a/（x-1）

f'(x)>0, x>a 2+1

The domain of f(x) is (1, positive infinity).

So f(x) decreases on (1, a 2+1) and increases on [a 2+1, positive infinity).

3. Let g(x)=f(x)-(5 8+ln2)=x -ax-aln(x-1)-5 8-ln2

According to the title, the image of g(x) has no common point with the x-axis.

g‘（x）=2x-a-a/（x-1）

When g'(x)=0, x=a 2+1 is solved

So just g(a 2+1)>0.

a/2+1）²-a（a/2+1）-aln（a/2）-5/8-ln2>0

A 4-aln(a 2)+3 8-ln2>0 takes a=1, then the above equation can be reduced to -1 4+ln2+3 8-ln2>0, that is, 1 8>0, which is obviously true.

So there is a = 1 that satisfies the requirement.

f'(x)=2x-a -a/(x-1)=x(2x-2-a)/(x-1) ，x>1

1) When a=1, f'(x)=x(2x-3) (x-1), let f'(x)=0, resulting in x=3 2

When x>3 2, f'(x)>0, f(x) is the increasing function, and when 11, so 2x-2-a>0, thus f'(x)=x(2x-2-a) (x-1)>0, and the increasing interval of f(x) is (1,+

When a>0, let f'(x)>0, the solution is x>1+a 2, the increasing interval is (1+a 2,+ the same reason, and the subtracting interval is (1,1+a 2).

3) When a=1, the minimum value of f(x) is 3 4 +ln2>5 8+ln2, so that y=f(x) has no intersection with y=5 8+ln2. That is, if there is a=1, the condition is satisfied.

1f'(x)=3x^2+2ax-1

f'(2/3)=4/3+4/3a-1=a ==> a=-1

2)f'(x)=3x^2-2x-1 =3(x-1)(x+1/3)

f'(x) >0 ==> x<-1 3, or x>1

f'(x)<0 ==> -1/31,f'x) >0, f(x) increment interval (1,+

f(x) minimum = f(1) = 1

2) If there is at least a little x0 in the interval [1,e], so that f(x0)<0 is true.

then x [1,e], f(x)min<0

f'（x)=-1/x^2+a/x =(ax-1)/x^2=a(x-1/a)

a<0, f'(x)<0 is constant, f(x) is decreasing, f(x)min=f(e)=1 e+a

1/e+a<0 ==>a<-1/e

0<1 a 1, a 1, x [1, e], f'(x) >0, f(x) increment, f(x) min = f(1) = 1, which is not in place.

1<1/a0,f(x)min=f(1/a)=a-alna=a(1-lna) >0

1 a e is 00, which is not on topic.

To sum up, a<-1 e

Is there a distribution to make it??

After finding the derivative, c is gone, so the equation about a is obtained, and a can be found, and if you ask the second question, c does not know that it does not affect the monotonicity, so you can directly find the derivative, and if you ask the third question, after the derivative, make the derivative greater than 0 and hold up on [-3,2].

I'll leave the second one.

1) d = (0, + infinity).

f'(x)=-1/x^2+1/x=0

Get x = 101 when f'(x)>0

The minimum value is f(1) when x=1

The monotonic interval (0,1) minus (1,+infinite) increases (2), that is, the minimum value of 1<=x<=e is required to be less than 0

f‘（x)=1/x(a-1/x)

Discuss the sign of a-1 x, find the corresponding minimum value and make it less than 0, that is, when 1 a<1 e is monotonically reduced, the minimum value is 1 e+a<0a<-1 e

A>1 There is no solution.

1 e in combination with a<-1 e