Let s take a look at the monotonic interval of the function f x x a x a 0 ,x 0, .

Find the monotonic interval of the function f(x)=x+a x(a>0),x(0,+.

Solution: Let 00, (x +x) 2+ a>0, x x >0, so the sign depends on the factor [(x +x) 2- a].

symbols. When 0 a, i.e., there is (x +x) 2- a>0 then f(x)-f(x) > 0, that is, in the interval (a,+) f(x) increases monotonically.

Seek guidance first, and get f'(x)=a-b/(x^2)

Let the derivative equal to 0 yields: a-b (x 2) = 0

The solution is x1 = (b a), x2 = - (b a) these two are the critical points.

When monotonically incrementing, f'(x)>0 is constant, and the solution is that x belongs to (- b a) and (b a),+

When monotonically decreasing, f'(x)<0 is constant, and the solution is that x belongs to - b a), 0) and (0, (b a).

It doesn't matter if the monotonic interval opens and closes.

x>0.

x=t (b a) substitution.

then a=b*t 2 x 2

fx=ax+b/x=bt^2/x+b/x=bt^2/(t√（b/a) )b/(t√（b/a) )

t√（ab)+√ab)/t

ab)(t+1/t)

t+1/t>=2√(t*1/t)=2

t=1 when the equal sign holds. That is, fx>=2 (ab)t=1 x= (b a) takes the minima, and on either side of the minima, they are monotonic functions, and the function is less than 0, which is similar.

Talk about the function and laugh about it.

f'Waiter guess (x) = 1-a x 2

Order f'(x)>0

x> a or x<- land type a

This is a monotonically increasing interval.

Such a solution would be a mistake of principle.

It is a mistake to use the monotonicity of the compound function as "increase and increase, increase and decrease to decrease, and decrease and increase" here.

The expression for the composite function is:

y=f(u),u=g(x), i.e. y=f(g(x)) and the expression here is.

y=f（x）g（x）

Such a function is not a composite function and cannot be applied to the above law.

Even if f(x) and g(x) are both increment functions, then y is right; If f(x) and g(x) increase and decrease, how can you determine whether the final loss is a stupid increase or a decrease? Similarly, if f and g are monotonically increasing or decreasing in y=f(x)+g(x), then the whole is also monotonically increasing or decreasing. But if it increases or decreases, it cannot be determined.

So this solution is wrong. The best solution to this problem is the derivative method.

y=x when the monotonic increase interval ,0].

The monotonic reduction interval is [0,-

y=e^ax

Evergrande is at zero. x∈r

is an increment function. So a> muffled implicit 0, the function f(x)=x 2*e ax. When monotonically increasing the destruction interval , 0].

The monotonic reduction interval is [0,-

The answer to this question is:

Solution: f(x)=3x -2mx-1

Let f (x) = o

Solution x = 3

or x = 3 when x or x > three monotonically increasing successively.

When 3 is monotonically decreasing.

(x)=x^2+ax-4 (|x|>=2) (1)f(x)=-x^2+ax+4 (|x|<2) (2)1.Since a>=0, so f(x) is monotonically increasing at x>=2 If 04, then f(x) is monotonically decreasing at x<=-a 2 and increasing monotonically at -a 2<=x<=-2.

If a>=4, then f(x) increases monotonically at -2<=x<=2 If 0<=a<4, then f(x) at -2<=x4, the single increase interval is [-a 2, + the single subtraction interval is (- a 2],

Solution: f(|x|)=x²+2|x|+3

1) x > mu group 0

f(x)=x²+2x+3

axis of symmetry x = -1

Therefore, it is absolutely resistant to wax, and it is on (0,6) and slips and increases singlely.

2)x<0

f(x)=x²-2x+3

axis of symmetry x = -1

So, decreasing on (-4,0).

(3) What does it mean.

f(x)=(x+1) +2 when a=1

x│∈[0，6]

In the interval x [0,6], f(x)=(x+1) +2 monotonically increases the closure of the socks.

When x [-4,0] is a monotonically decreasing interval in f(x)=(x+1) +2.

The limbs are good, and the file is messy.

1) The monotonic increase interval is <>

The monotonic reduction interval is <>

2) When 0 knows that the analytic formula of the function is to find the monotonic interval, the essence is to find the solution interval of f (x)>0, f (x) < 0, and pay attention to defining the domain;

The monotonicity of f(x) on [1,2] is studied first, and then the maximum value is determined whether it is an endpoint value or an extreme value.

Since the analytic formula contains parameter a, it is necessary to classify and discuss parameter a in specification (1)f (x) <

a(x>0) (1 point).

When a 0, f (x) <

a 0, i.e., the monotonic increase interval of the function f(x) is (0, 3 points) when a>0, let f(x) <

a 0, get x <>

When 0<>

f(x) <

0 when x > <

f(x) <

0, so the monotonic increase interval of the function f(x) is <>

The monotonic reduction interval is <>

6 points) 2) When <>

1, i.e., a 1, the function f(x) is a subtraction function over the interval [1,2], so the minimum value of f(x) is f(2) ln2 2a(8 points).

When <> 2, it is 0<>

, the function f(x) is an increasing function over the interval [1,2], so the minimum value of f(x) is f(1) a(10 points).

When 1 < <

2, i.e. <>

is the increment function on and is <> in the interval

is a subtraction function, and f(2) f(1) ln2 a, so when <>

(1)f(x)=ax^3-3x, f’(x)=3ax^2-3=3(ax^2-1).Because a 0, so f'(x) 0, so f(x) is monotonically decreasing on r.

2)a.When a 0, the minimum value is f(2)=8a-6<0 so the minimum value cannot =4, so a 0, which is not on topic.

b.When a>0, let f'(x)=0, get 3ax 2-3=0, x= a a

,a a),(a a,+ is a monotonically increasing interval, and (-a a, a a) is a monotonically decreasing interval.

a) When a a 1, i.e., a 1, the interval [1,2] is an increasing interval, so the minimum value = f(1) = a-3 = 4, where a = 7

2) When 1< a a <2, i.e. 1 4 (c) When a a 2, i.e. 0 so the minimum value = f(2) = 8a-6 = 4, a = 5 4 no (0, 1 4], in summary, a = 7

1)f'(x)=3ax²-3

When a<=0, f'(x)<0, the function is monotonically reduced on r.

2) If a<=0, then monotonically decrease on [1,2], and the minimum value is f(2)=8a-6=4, obtain: a=5 4, contradictory;

If a>0, then there is a maximum point x=-1 a, and the minimum point x=1 a If a>1, then it increases monotonically in [1,2], and the minimum value is f(1)=a-3=4, and obtains: a=7, which is in line with;

If 0 if 1 4=