If the function f x 1 3ax3 bx c x r is an odd function on r, and the function takes an extreme value

Since the function f(x)=(a 3) x 3+bx+c is an odd function, then f(0)=c=0

Because f(x) takes the extreme value at x=1, and f'(x)=ax 2+b, then there must be f'(1)=a+b=0, i.e. a=-b.

And because the extreme value f(1)=(a3)+b=1

So, a=-3 2, b=3 2

From this we see the function f(x)=-(1 2)x 3+(3 2)x

f(x)=1 3ax +bx+c(x r) does not contain even terms, it is an odd function on r, so the constant term c=0

f′(x)=ax²+b

Extreme values are obtained at x=1.

f′(1)=a+b=0

b=-af(x)=1/3ax³-ax

f(1)=-1

1/3a-a=-1

a=3/2，b=-3/2，c=0

f(x)=1/3ax³-ax

f(e^x)=1/3ae^(3x)-ae^x

f′(x)=ae^(3x)-ae^x=ae^x=ae^x(e^x+1)(e^x-1)

a＞0，ae^x(e^x+1)＞0

f(e x) decreases monotonically when x 0, and increases monotonically when x 0.

The minimum value of f(e x) = f(e 0) = f(1) = 1 3a - a = -2 3a

Under the condition of (1), f(x) = 1 2x -3 2x

f′(x)=3/2(x+1)(x-1)

The maximum value f(-1) = -1 2 + 3 2 = 1

The minimum f(1) = 1 2-3 2 = -1

If y=k and y=f(x) have three different intersections, then the k value must be between the maximum and minimum (and not including the extremum), i.e.

f(1)＜k＜f(-1)

1＜k＜1

f(x)=[x+1)(x+a)] Hand round x

x^2+(a+1)x+a]/x

x+a/x+(a+1)

To be the odd letter Bisen collapse number spring resistance, constant term = 0

a+1=0a=-1

First of all, analyze the condition of the problem 1, f(x) is an odd function So: f(-x) = -f(x).

2. When x=1, the extreme value of the function is 2

Then: f(1)=a+c+d=2 *

You can know that f(-1)=-a+c)+d=-2 3, according to its extreme value condition, the derivative of the original function can be found, so the value of the function is 0 when x=1: f'(1)=3a+c=0

Then the value of ab c can be found by concatenating the three formulas of *, and the analytic formula of the original function is found.

The later verification will give you an idea!

It is not difficult to determine the monotonicity (monotonic decreasing) of the function on the closed interval of 1 to 1, so it is only necessary to prove that the maximum difference between the values of the function is less than or equal to 4, that is, when x1 and x2 go to the two endpoints of the interval and then find the function, it can be compared and proved.

Proof: Since the function is an odd function, f(x)=f(-x), there is ax 3+cx+d=a(-x) 3-cx+d, so d=0

Since the function takes the minimum value -2 at x=1, there is f'(1)=3a+c=0, f(1)=a+c=-2, so a=1, c=-3

Therefore f(x)=x 3-3x.

Then the derivative of f(x) shows that f(x) decreases at (-1,1), so that the maximum value of f(x) is: f(-1)=2.

f(x1)-f(x2）|

The odd function --- d=0, and when x=1, the extreme value of -2--- a=1, c=-3 is obtained; f(x）=x^3-3x。。。I'll do it later.

f(x)=ax 3+bx 2+cx(a≠0) is the odd function defined on r, b=0,f'(x)=3ax 2+c,x=-1, the function takes the extreme value 1,f'(-1)=3a+c=0,f(-1)=-a-c=1.

The solution yields a=1 2, c=-3

f'(x)=(3/2)(x+1)(x-1),-1|f(x1)-f(x2)|< = f(-1)-f(1)=1-(-1)=2<=s, the minimum value of s = 2

The function f(x) is an odd function, f(-x)=-f(x) so -[ax 3+bx 2+cx+d]=a(-x) 3+b(-x) 2-cx+d

So b=0, d=0

So f=ax 3+cx

f'=3ax^2+c

When x 1, f(x) has a minimum value of -3 2.

So x=1 is f'=0 for a root, so 3a+c=0f(1)=a+c=-3 2

The simultaneous equation yields: a=3 4, c=-9 4

f(x)=3/4x^3-9/4x

f'(x)=9 4x 2-9 4=9 4(x 2-1) so when -1=f(1)=3 4-9 4=-3 2 so, |f(x1)-f(x2)|<3/2-(-3/2)=3

1) Solution: From f(-x)=-f(x), we can know that b=d=0, and f(1) is the minima, therefore.

f(1)=-2/3,f'(1)=0.The solution yields a=1 3, c=-1i.e. f(x)=1 3x 3-x 2) proves: f'(x)=x~2-1<=0,(x€［-1,1］

Next

Because f(x) is an odd function, b=0, and because x=1 takes the minimum value, the derivative of f(x) at x=1 is 0, i.e., f'(x)=3ax 2+c, bring x=1 in, and get 3a+c=0

For odd functions, take the minimum value at x=1 and the maximum value at x=-1.

f(1)=a+c+d=-2/3 ②

f(-1)=-a-c+d=2/3 ③

It's time to figure it out.

Needless to say, the second question is at (-1,1) the maximum value is 2 3 and the minimum value is -2 3

f（x)'=3ax 2+2bx+c, when x=1, f(x)mix=-2 3. -2 3=3a+2b+c, there is a formula after it, I forgot. Excuse me, but let's work out a, b, c

(1) Odd function Explain f(0)=0, and get d=

f'(x)=3ax^2+2bx+c,f'(1)=0, we get 3a+2b+c=0

The minimum is -2 3, giving f(1) = -2 3i.e.: a+b+c+d=-2 3

Bringing in calculates: a=1 3, c=-1

f(x)=x^3/3-x

2) Obviously, as long as the difference between the maximum and minimum values on [-1,1] is proved to be satisfied, then everything else is true!

f'(x)=x 2-1 is subtracted when on [-1,1]. So when f(-1) there is a maximum value and f(1) is the minimum value.

So there is: |f(x1)-f(x2)|<=|f(-1)-f(1)|=|2/3+2/3|=4/3

Odd functions have a constant term of 0, so d=0

f(x)=ax^3+cx

f'(x)=3ax^2+c

f'(1)=3a+c=0

f(1)=a+c=-2

Solution: a=1, c=-3

f(x)=x^3-3x

1) f'(x)=3x 2-3=3(x+1)(x-1)=0, and the extreme point is -1, 1

x<-1 or x>1, monotonically increased.

To find the derivative, substituting x=1 for f-derivative = 0, substituting x=1, f=-2 because it is an odd function, for x=0, f=0

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

It is known that there is f(0)=d=0 and f'(1)=3a+c=0 and f(1)=a+c+d=-2

The solution yields a=1, c=-3, and d=0

So f(x)=x -3x

2)f(x)=x³-3x

f'(x) = 3 x 2-3 = 3 (x + 1) (x - 1) by f'(x) >0, and the solution is x<-1 or x>1

f'(x) <0, the solution is -1f'(-1)=f'(1)=0

So the single increase interval of f(x) is (- 1), and the (1,+ single decrease interval is (-1,1).

f(x) has a maximum value f(-1)=2, and a minimum value f(1)=-2, I hope it can help you!