The function y x3 2x2 x 3 is known, and the extreme points and monotonic intervals of this function

y'=3x^2-4x+1x=1，y'=0y extremum=3

1 For example: Ling F'(x)=3x²-3=0

x=±1x<-1,x>1，f'(x) >0, incremental.

1 So. Increase the intervals (- 1) and (1, +

Minus interval (-1,1).

The maximum value is f(-1)=0

The minimum f(1) = -4

Extremum Points: Function images.

The abscissa of the upper maximum or minimum point in a sub-interval of the segment.

Extreme points appear at the stationary point of the function.

a point with a derivative of 0) or a non-derivative point (a derivative function.

does not exist, you can also get an extreme value, in which case the station does not exist).

If f(a) is the maximum or minimum of the function f(x), then a is the extreme point of the function f(x), and the maximum point and the minimum point are collectively called the extreme point. The extreme point is the abscissa of the upper maximum or minimum point in a sub-interval of the function image. Extreme points occur at the stationary point of the function (the point with a 0 derivative) or at the non-derivable point (the derivative function does not exist, and the extreme value can also be obtained, in which case the stationary point does not exist).

The zero point of the derivative function is the extreme point of the original function, y'=3x 2-4x=x(3x-4), x=0 or 4 3, y'=0, then x=0 or 4 3 is the extreme point of the original function. x<0 or x>4 3, y'> 0,04 3, monotonically decreasing interval of 0

y'=3x^2-4x+1

x=1,y'=0y extremum=3

x=3,y'=0, y extremum=21

x<1,y'>0 ,x>3,y'>0 monotonically incrementing.

1

Wow someone asked so quickly.

The details are as follows:y''(1) <0 x=-1 is the maximum point, and the maximum value = y(-1)=5y''(3)>0 x=+3 is the minimum point, and the minimum value =y(-1)=-27x (-1) (3,+ is the monotonically increasing interval.

x (-1,3) is a monotonically decreasing interval.

Monotonicity of functions:Let the domain of the function f(x) be d, and the interval i is contained in d. If for any two points x1 and x2 on the interval and when x1 is for any two points x1 and x2 on the interval i, when x1f(x2), then the function f(x) is said to be monotonically decreasing on the interval i, and the functions of monotonically increasing and monotonically decreasing are collectively called monotonic functions.

The derivative of y is 3x 2-6x -9=3(x-3)(x+1), when the derivative of y is greater than zero, i.e., x>3, x<-1 monotonically increases, and vice versa, the monotonically decreasing extrema is obtained at x=3 and x=-1.

When x = 3, y is the minimum value of -27

When x = -1, y is maximum, and the value is 5

Derivative, define the domain as r, and the derivative function y'=3x 2-6x-9=(3x+3)(x-3) 0, which solves the unary quadratic inequality.

It can be seen that the monotony is reduced at [-1,3]. (Both ends of the interval can be opened or closed) (negative infinity, -1) and (3, positive infinity) monotonically increased.

Extremums bring -1 and 3 in.

The maximum value is 5, and the minimum value is -17

Its derivative is y=3x 2-6x-9 can be reduced to y=3(x-1) 2-12, and when x=1, y has a minimum value of -12,

Routine operations: find the derivative, find the zero point of the derivative function, run the judgment column and fight the table, and write the answer.

For reference, please smile.

The idea is clear, concise and fluent.

y'=3x^2-4x+1

x=1,y'=0y extremum=3

x=3,y'=0, y extremum=21

x0 ,x>3,y'>0 Single jujube dust is increasing in imitation.

1

The solution is derived from y=x3-3x2+10 to obtain y'=3x 2-6x require'=0 to get x=0 or x=2 when x belongs to (negative infinity, 0), f'Acacia (x) 0 When x belongs to (0,2), f'(x) 0 When x belongs to the (0, positive infinity) ridge, f'(x) 0, so the increasing interval of the function is (negative infinity, 0) and (0, positive infinity) before the decrease is (0, 2)...

y'=3x^2-8x-3

3x+1)(x-3)=0

When x -1 3 or x > judgment 3, y has a single bend tone increasing buried punch range.

When -1 3

Derivative, get y'=3x²+6x-24

When y'=0 gives x=-4 or 2

Therefore, the monotonically decreasing interval of this function is [-4,2], and the monotonic increasing interval is (-4] and [2,+).

The maximum value is 81 and the minimum value is -27

y＝x³＋3x²－24x＋1

y'3x Land Type 6x 24

Order y'0 gets x early mountain guess 4, x 2 is easy to get increasing intervals (4) and (2, decreasing intervals (4, 2), x 4 takes the maximum value of 81, x 2 takes the minimum value of 27.

y＝x^3＋3x^2－1,∴y′＝3x^2＋6x,y″＝6x＋6.Let y 0, get: 3x 2 leak 6x 晌運shed 0, x(x 2) 0, x 0, or x 2

Obviously, when x 0 and y 6 0, then the function feast has a minimum value of 1When x 2, y 6 ( 2) 6 6 0, the function has at this point.

Requiring the monotonic interval, extremum, and extremum point of the function $y=3x 4+2x 3-1$, you can first find its derivative:

y'=12x^3+6x^2$$

Order y'=0$, get:

x^2(2x+1)=0$$

The solution is $x 1=0$ and $x 2=- frac$. Bringing these two solutions back into the original function gives us the corresponding $y$ values, which are $y 1=-1$ and $y 2= frac$, respectively.

Therefore, the extreme points of $y$ are $(0,-1)$ and $(frac, frac)$.

Next, the monotonicity of the derivative can be used to find the monotonic interval of the function.

When $x<, $y'<0$,$y$ monotonically decreasing; When $,y$ increases monotonically; When $x>0$, $y'>0$, $y$ is monotonically incremented.

In summary, the monotonic intervals of function $y=3x 4+2x 3-1$ are $(infty,- frac)$ and $(0, infty)$, the maximum point is $(frac, frac)$, and the minimum point is $(0,-1)$.

y＝x^3＋3x^2－1,∴y′＝3x^2＋6x,y″＝6x＋6.

Let y 0, get: 3x 2 6x 0, x(x 2) 0, x 0, or x 2

Obviously, when x 0 and y 6 0, the function has a minimum value of 1

When x 2, y 6 ( 2) 6 6 0, then the function has a maximum value of ( 2) 3 3 ( 2) 2 laughing mountain 1 3

Naturally, the increasing interval of the function is (1) (0 and decreasing is (1,0),3,