It is known that f x 1 3ax bx 2 is a function defined on R, and f x is a subtractive function on

Solution: Derivative: f'(x)=ax2+b, the derivative is a continuous function.

Because on (0,1) is the subtraction function at (1,+ is the increasing function (the derivative of the subtraction function is less than 0, and the derivative of the increasing function is greater than 0), and the derivative is 0 when x=1

Get f'(1)=a+b=0

Again: the tangent slope at x=0 is the derivative at x=0, f'(0)=b,y=x+2 has a slope of 1 (the derivative is 1, so the slope is 1), that is, it is at an angle of 45 degrees to the x-axis, so the slope of the tangent is 135 degrees to the x-axis (just make a diagram to understand), and get b=-1

Or there can be: the product of the slopes of two lines perpendicular to each other is -1, so we get b=-1 and bring b=-1 into a+b=0 to get a=1

So f(x)=1 3x +x+2

Question: The function defined on r f(x)=1 3ax3+bx2+cx+2 satisfies the following conditions at the same time:

On (0,1) is a function of time and on (1,+infinity) is an increasing function.

The guide is an even function.

The tangent at x=0 is perpendicular to the straight line y=x+2.

Question:1Find the analytic expression of the function f(x).

2.Let g(x)=(1 3x3-f(x))ex, find the minimum value of the function g(x) on [m,m+1].

Answer: f'(x)=ax +2bx+c, due to f'(x) is an even function, so b=0,f'(x)=ax²+c

f(x) is a subtracting function on (0,1) and an increasing function on (1,+infinity), so f'(1)=0, i.e. a+c=0

The tangent of f(x) at x=0 is perpendicular to the straight line y=x+2, so f'(0)=c=-1, thus a=1

So f(x)=x 3 -x+2

g'(x)=e x+(x-2)e x=(x-1)e x, it is easy to know that g(x) is a subtraction function on (- 1] and an increasing function on [1,+.

1) When 0 m 1, 1 [m, m+1], [g(x)]min=g(1)=-e;

2) When m<0, g(x) is a subtraction function on [m,m+1], [g(x)]min=g(m+1)=(m-1)e (m+1);

3) When m>1, g(x) is an increasing function on [m,m+1], [g(x)]min=g(m)=(m-2)e m

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The answer is definitely wrong, and it is not true when the book is at least a=0.

Subtract the function. So big is f'(x)=3ax +6x-1 0 constant established.

Obviously a=0,f'(x)=6x-1 is not true.

a≠0, then the quadratic function opening is held down to the state.

a<0 and discriminant 36+12a 0

a≤-3

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I'll call you when I ask.

1) Find the derivative function f'(x)=3ax +2bx+c has opposite monotonicity in the intervals [-1,0] and [0,2], drawing the derivative image.

Then x=0 must be f'(x) of a zero.

i.e. f' (0)=0 ∴c=0

3) When f'(x)=0.

i.e. 3ax +2bx=0

x1=0 x2=-2b/3a

f(x) has opposite monotonicity in the intervals [0,2] and [4,5] 2 -2b 3a 4

6≤a/b≤-3

Because f(x) knows that one zero point x=2 lets the other two zero points a(m,0) c(n,0).

Then 丨ac丨=丨m-n丨= (m-n) = [(m+n) -4mn] is used by the undetermined coefficient method.

f(x)=a(x-m)(x-2)(x-n)=ax³-a(m+n+2)x²+a(2n+mn+2m)x-2amn=ax³+bx²+cx+d

b= -a(m+n+2) ①

c=a(2n+mn+2m)=0 ②

d= -2amn ③

f(2)=8a+4b+d=0 ④

The simultaneous solution yields m+n= (-b a) -2

mn=4+ (2b/a)

ac=√[(m+n)²-4mn]

[(b/a -2)²-4(4+ 2b/a)]=√[(b/a)²-4b/a)+4-16]=√[(b/a -2)²-16]

6≤b/a≤-3

Think of it as a quadratic function with an axis of symmetry of 2

b a=-6 has a maximum value of 4 and 3b a=-3 has a minimum value of 3

ac belongs to the interval [3,4 3].

I figured it out.

2) Pushed by (3).

6≤b/a≤-3

The derivative function represents the change in the slope of the function.

i.e. if m(x0,y0) is present so that the slope is 3b

That is, the derivative function f at this time' (x0)=3b

i.e. 3a(x0) 2bx0=3b

That is, 3a(x0) 2bx0-3b=0 has a solution, and it may be assumed that this point m exists

That is, the equation has a solution.

then the 0 of the equation

Equation =4b -4 3a (3b)=4b +36ab=4ab[(b a)+9].

6≤b/a≤-3

a,b heterogeneous sign then ab 0 and (b a )+9 0 0

Contradiction, so there is no m, making its tangent 3b

If you don't understand, please feel free to ask.

1) Find the derivative function f'(x)=3ax +2bx+c has opposite monotonicity in the intervals [-1,0] and [0,2], and x=0 must be f according to the derivative function'(x) of a zero.

i.e. f' (0)=0 ∴c=0

2) According to f(x)=ax +bx +cx+d is a function defined on r, it must be deduced that this function is a continuous function, and 0 to 2 and 2 to 4 are the same monotonic intervals, so it is further deduced that 0 and 4 are the two poles of the original function, and are the two roots of the derivative function, so bring in the derivative function to find a=-b 6, so f'(x)=-bx 2+2bx=3b solves x -4x+6=0 equation without solution, so there is no such point m

3) When f'(x)=0, i.e., 3ax +2bx=0 gives x1=0 x2=-2b 3a

f(x) has opposite monotonicity in the intervals [0,2] and [4,5] 2 -2b 3a 4 6 a b -3

And f(x) is known to have one zero point x=2, so the other two zero points can be set to a(m,0) c(n,0).

Then 丨ac丨=丨m-n丨= (m-n) = [(m+n) -4mn] can be deduced from the undetermined coefficient method.

f(x)=a(x-m)(x-2)(x-n) =ax³-a(m+n+2)x²+a(2n+mn+2m)x-2amn =ax³+bx²+cx+d

b= -a(m+n+2) ①c=a(2n+mn+2m)=0 ② d= -2amn ③f(2)=8a+4b+d=0 ④

Joining these four sets of equations gives m+n= (-b a) -2 mn=4+ (2b a).

ac=√[(m+n)²-4mn] =√[(b/a -2)²-4(4+ 2b/a)] =√[(b/a)²-4b/a)+4-16]=√[(b/a -2)²-16]

-6 b a -3 change order b a=t then ac= [(t -2) -16].

Then -6 t -3 can be regarded as a quadratic function with an opening upwards and an axis of symmetry of 2 so that it decreases monotonically on [-6,-3].

t=-6 has a maximum value of 4 3

t=-3 has a minimum value of 3

ac belongs to the interval [3,4 3].

When x>=0, f(x)=-x 2+1 is the subtracting ridge function.

And when x=0, f(x)max=f(0)=-0 2+1=1f(x) is a subtraction function on r.

When x<0, f(x)>f(0)=1

x+3a>1

3a>x+1

When x takes the most coincidental lead value 0 from the x-negative semi-axis, there are: 3a>1a>1 3

The value range of a: (1 3,+.)

f'(x)=(x²+ax-2a²+3a+2x+a)e^x=[x²+(a+2)x-2a(a-2)]e^x=(x+2a)(x+2-a)e^x

by f'(x)=0, we get x1=-2a, x2=a-2 because a≠2 3, then x1≠x2

Therefore, x1 and x2 are extreme points, f(x1=3ae (-2a), f(x2)=(-3a+4)e (a-2).

1) When a>2 3, x2 > x1

The monotonic increase interval is: xx2; The monotonic reduction interval is (x1, x2) and the maximum is f(x1)=3ae (-2a).

The minimum is f(x2)=(-3a+4)e(a-2)2) when a<2 3, x1>x2

The monotonic increase interval is: xx1; The monotonic reduction interval is (x2, x1), the maximum is f(x2)=(-3a+4)e (a-2), and the minimum is f(x1)=3ae (-2a).

Solution: 1)f'(x)=2x+2-1 (2x 2), obviously x [1, when f'(x)>0, f(x) is an increasing function.

f(x)min=f(1)=7/2

2) f(x)>0 constant establishment.

x 2+2x+a x>0 to any x belongs to [1, positive infinity) constant.

That is, x 3+2x 2+a>0 to any x belongs to [1, positive infinity) constantly.

i.e. a>-(x 3+2x 2) so that g(x)=-x 3+2x 2)g'(x)=-3x 2-4x

g'(x) is a subtractive function on x [1, .

a>g（1）=-3

i.e. a (-3,

1. f(x)=(x²+2x+,x∈[1,+(x²+2x+1)

x+1) because (x+1) x>=0

So the minimum value of the function f(x) is .

x∈[1,+

x²+2x+1)+a-1/x

x+1)²+a-1/x

Because (x+1) +a-1 x>0

So A-1>0

a>1

f(x)=x+a/x+2 （x>=1)

When x = a, f(x) takes the small value.

f(√a)=2√a+2=√2+2

Among them: Nike function f(x)=x+k x.

For y=x+k x, it is called the tick function Also known as the Nike function (the symbol of the image item Nike), when the question is asked, k>0 is generally set, which is an odd function!

In (- k), (k, + are the increasing functions.

At (-k,0), (0, k) is a subtractive function!!

The value range is: (-2 k), (2 k, +.)