Urgently ask a high school function monotonicity problem with the help of the master

Since f(x) is an increment function, if you want it to increase, x must also be incremented.

Then you ask for the increment interval of x. This x is equivalent to your (-x 2+5x+6), which is the incrementing interval of (-x 2+5x+6).

There are two roots of x 2+5x+6=0, one is -1 and the other is 6. For the equation -x 2 + 5x + 6 = 0

Finding the derivative, we get -2x+5=0, where x= is its maximum value. You can also draw a graph to come to the same conclusion. And note that the function is defined on positive infinity, and negative values are meaningless.

So the value is between -1 and 6 (looking at the graph, only the difference between -1 and 6 is above the x-axis). Its increment interval is -1 to. So -1 doesn't know if it's detailed enough.

By definition, when 00, the solution gives -10

x+1)(x-6)<0

The monotonic interval is [-1,6].

Since the domain of x is +infinity, starting with -x2+5x+6>0 gives -10 to get x> which can also be equal to oh, and because x<6 its monotonic increase interval is

Solve with a coincidence function:

Since f(x) is incremented over (0, ), the domain of x is defined as (0, ).

Then take the ones in parentheses out of them as g(x) and think of the one you're asking for as a composite function of f(x) and g(x).

That is, let g(x)=-x 2+5x+6=-(x-5 2) 2+49 4 (2 is squared).

and x (0.∞)

So g(x) increases at (0,5 2) and (5 2,+ decreases.)

Because f(-x 2+5x+6) is a composite function of f(x) and g(x), and f(x) increases at (0, ), g(x) increases at (0,5 2), and (5 2,+ decreases).

So f(-x 2+5x+6) increases at (0,5 2) and (5 2,+ subtracts).

ps: the same increase and different decrease, that is, f(x) and g(x) are both increasing, then the composite function is increasing; If one of the two increases and one subtracts, their composite function is subtraction.

f(x) is incremented on (0,+&. If there is f(a), then let t=-x 2+5x+6, t must be in the defined domain (0,+&, so find x in.

1< x < 6 to have monotonicity, increasing at -1; In< x<6, decreasing.

Note, x2 is the square of x].

From -x2+5x+6>0, -10 yields y=-x2+5x+6 with an increase interval of -1< x <, let -2x+5<0 get the subtract interval of y=-x2+5x+6

Hehe, because the definition domain of x is +infinity, starting with -x2+5x+6>0 gets -10 to get x> can also be equal to oh, the same increase is increasing, and because x<6 so its monotonous increase interval is

x^2+5x+6>0

x+1)(x-6)<0

When -1 x is used, y=-x 2+5x+6 is monotonically incrementing, which is the same as f(x).

When x (-1,,f(-x2+5x+6) is monotonically decreasing, y=-x 2+5x+6 is monotonically decreasing, and f(x) is an increasing function.

When x(,6), f(-x2+5x+6) is monotonically decreasing.

Knowing that the function f(x) is an increasing function defined on positive infinity, try to find that the monotonic interval of the function f(-x2+5x+6) is —(the square of x).

Solution: f(x)=-(x-5 2) 2+49 4 When 5 2<=x, it decreases monotonically.

When x<= 5 2, it is monotonically incremented.

This function is a subtractive function, which means that as x increases the track and the accompanying rock increases, the corresponding y value gradually decreases. Mess up.

So in f(x)>f(2-x), the y value on the left is larger than the y value on the right, which means that the x value on the left is smaller than the x value on the right.

This gives x<2-x.

Khan: Let's look at the definition domain first, which has already been given in the question: x>1

then x1>x2>1 x1, x2 is an arbitrary real number.

Because: x1>x2>0

So: x1 2> x2 2, then x1, x2>1 So: x1 2-1> x2 2-1>0

So: sqrt(x1 2-1)>sqrt(x2 2-1) leads to the conclusion: (1, positive infinity).

If x1>x2 then f(x1) > f(x2).

That is, f(x) increases monotonically in this interval

This is the basic way to judge the monotony of the function, take x1>x2 in the definition domain, and then compare f(x1), f(x2), if the comparison is not clear, divide the definition domain into several small segments and then compare.

Method 1: Derivative method.

Derivative of f is obtained'=x (x 2-1) when x > 1, x 2-1>0

So f'>0

So the monotonically increasing method of f(x) on (1, positive infinity) is the definition method.

x1>x2>1

f(x2)-f(x1)

(x2^2-1)-√x1^2-1)

√(x2^2-1)+√x1^2-1))/(√(x2^2-1)-√x1^2-1))*x2^2-1)+√x1^2-1))

√(x2^2-1)+√x1^2-1))/（x2^2-x1^2）>0

The same can be proved to be an increasing function.

Derivative, f(x) is obtained'=x (x 2-1) when x > 1, x 2-1>0

So f(x).'>0

So f(x) increases monotonically over (1, positive infinity).

Monotonic increasing function, the derivative of the function, from the defined domain, can be known to be greater than zero.

Method 1: Derivative method.

Seek a derivative of f, and know f'=x channel (x 2-1) when x > 1

time, x 2-1>0

So f'>0

So f(x) is monotonically incremented on (1, positive infinity).

Method 2: Definition.

x1>x2>1

f(x2)-f(x1)

x2^2-1)-√x1^2-1)

(x2^2-1)+√x1^2-1))/x2^2-1)-√x1^2-1))*x2^2-1)+√x1^2-1))

(x2 2-1)+ x1 2-1)) x2 2-x1 2) >0 can also belong to the same to prove to be an increasing function.

These two questions are about abstract functions, you can bring a specific function into as long as it meets the meaning of the question, you can know the nature of the function from the special, you know it in your heart, and you will not be too blind when proving.

Let's talk about the specific solution:

1) Shilling x=1 x, bring in the function to get: 2f(1 x)+f(x)=1 x Multiply both sides of the original formula by 2 and subtract it from this equation, try it yourself. After the analytic formula comes out, the parity can be judged according to the definition.

2) Let x=-y, bring in the original equation, and you can get the result you want.

I wonder if it's clear?