f X X 3 1 2X 2 bX c is known, and f x achieves an extreme value at x 1

Seek guidance first. '(x)=3x^2-x+b

When f'(x)=0, a root is 1, i.e., 3-1+b=0, b=-22Finish the zero point of the first inverse function, 3x 2-x-2=0 and push out x=1 or x=-2 3

When x》1, the derivative is greater than 0, so the function is incremented.

So f(2) is a quadratic function, and c belongs to negative infinity to -1 and 2 to positive infinity.

3.This needs to have a graph, and as you can see from this function, the image is first increasing and then decreasing and then increasing.

Just like in this diagram, c controls the downward panning of the image, so as long as there is only one intersection point with x once you reach that range.

Here's how to solve it. Since the two zeros of the derivative function are 1, -2 3, first remove cf(1) = f(-2 3) = 22 27

Remove c, the function is over (0,0).

You can probably draw an image based on this, and you can find out.

c less than -22 27 or greater than.

Derivation f'（x)=3x^2-x+b

1) 3-1+b=0 b=-2 when x=1

2)f'（x)=3x^2-x-2

Order f'(x)=3x 2-x-2 less than 0 solution -2 3x or x>1 monotonically increases.

f(x) in the interval [1,2] the minimum value is -3 2+c maximum value 2+ c23) when the maximum value is less than 0 or the minimum value is greater than 0, the curve y=f(x) has only one intersection point with the x-axis.

So the minimum value -3 2+c>0 or the maximum value 22 27+c<0 is solved to get 3 2c

1、f'(x)=3x²+2ax+b=0

Both x=1 and -2 have extremums.

So x=1 and -2 are the roots of the equation.

By the Vedic theorem.

2a/3=-(-2+1)=1

b/3=-2*1=-2

a=3/2,b=-6

2、f'(x)=3x²+3x-6

So -21 is an increment function, so x=1 has a minima.

So f(1)=0

1+3/2-6+c=0

c=7/2

f'The solution of (x)=3x 2+2ax+b=0 is 1, and -2 is substituted into:

3+2a+b=0

12-4a+b=0

a=,b=-6

f(x)=x^3+

f(1)=1+

f(-2)=-8+6+12+c=c+10

Therefore, min f(x)=f(1)=, x [-3,2] is c=

Quadratic functions.

x = -2 3, x = 1 is equal to 0

The opening is upward. So x<-2 3, x>1 is greater than 0

Others and so on.

Isn't there a lack of conditions for the topic?

The problem I saw was "If f(-1)=3 2, find the monotonic interval and extreme value of f(x)".

A should be -1 2 b should be -2 so f'(x)=3x 2-x-2 f'(x)>0 so x<-2 3,x>1,f'(x) >0 and f(x) increment.

f'(x)<0 So-2 3

Hello! The extremum is obtained first, then the derivative is 0

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

f'(1) =-3+2a-3=0, the solution is a=3, and f(2)=-8+4a-6+b=4a+b-14=312-14+b=3

b-2=3b=5So, b=3, b=5

I wish you progress in your studies o(oha!

If you don't understand it in the future, you can still hi me.

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

At x=1, the extreme value is obtained.

f'(1)=0

i.e. -3+2a-3=0

a=3f(2)=-8+4a-6+b=3

Get b=5

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

The derivative of the extreme point is 0

f'(-1)=3-2a+b=0

f'(2)=12+4a+b=0

a=-3/2,b=-6

f(x)=x^3-3/2x^2-6x+c

f'(x)=3x^2-3x-6=3(x^2-x-2)=3(x+1)(x-2)

f'(x)<0,-1 so f(x) decreases monotonically at (-1,2) and increases monotonically at (-1), (2,+ monotonically.

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

1+2=-2a/3, (1)*2=b/3a=-3/2,b=-6

2)f'(x)=3x2-(3 2)x-6<0 =>-10 =>x<-1 or x>2

Monotonic reduction interval (-1,2) of the function f(x).

Monotonic increase interval (-1] [2+) of the function f(x).

1、f'(x)=3x 2+2ax+b has two roots, 1 and 2 3, substituting 3+2a+b 0, 4 3 4a 3+b=0, and the solution gives a 1 2, b 2. Therefore f(x)=x 3 x 2 2 2x+c, the minimum f(1)=c 3 2, and the maximum f( 2 3)=22 27+c.

2. f(2)=2+c, f(x) increases on [1, 2 3], decreases on [2 3,1], and increases on [1,2], so the condition is equivalent to the maximum value of f(x) on [1,2] < 1 c, that is, 2+c<1 c. Solution.

c< 1 root number (2) or 0

c is the same algebra, but 22 7 is greater than 2, so you can determine whether 22 7+c is greater or 2+c is larger.