A few questions with high school functions, urgent 10

I'm a freshman in high school, and finding a derivative is finding a derivative function, and the derivative is the slope, and then, in fact, the basic knowledge of calculus is very simple, you can see it yourself, I will be in my third year of junior high school, and now I'll talk about the specific operation:'=(f(x+h)-f(x)) h=3 ((x+4)*(x+4)), which is obvious: at infinity x -4, f(x) is an increasing function; When infinitesimal x -4, f(x) is also an increasing function.

2.Let x1 and x2, and x1 x2, f(x1)-f(x2)=3(x1-x2) (x1+4) (x2+4), because x1-x2 0, then: when infinity x -4, f(x1)-f(x2) 0, f(x) is an increasing function; When infinitesimal x -4, f(x1)-f(x2) 0, f(x) is also an increasing function.

f(x)=(x+1) (x+4)=1-, with x=-4 as the axis, you can make its image, then the monotonicity is clear at a glance!

In fact, f(x)=(ax+b) (cx+d) can be done in this way, f(x)=(a c)+, with x=-d c as the axis, you can draw the image of the function, then the monotonicity is found!

If you learn calculus, it's too easy, so let's use the elementary method.

It can also be proved by the definition of monotonicity!

Damn, you should use calculus is something that every middle school student knows. Landlord, since you can think of it.

f(x)=(x+1) (x+1)+3=1+(x+1) 3 So why not use the following idea of subtraction? f(x)=(x=4-3) (x+4)=1-3 (x+4)You only need to consider the monotonicity of -1 (x+4), why bother to know the connection between x 3 and x 1?

Brother, let me tell you, the general way to find the monotonic interval of an elementary function is to find the derivative, the basic elementary function is derivable in its defined domain, first find the non-derivative, and then find the derivative = 0 point.

It can be judged by the positive and negative of the derivative function on the interval.

Differential calculus and integral calculus and calculus.

I can give the definition of the derivative, but you may not understand it, because the derivative is defined in terms of the limit of the function.

The simple derivative is: limδy δx, when δx approaches to 0.

The definition of indefinite integral is given in terms of derivative, simply put: the family of original functions of the derivative function is indefinite integral.

I'm a junior now, and I can't even say I know calculus!

Boss, can't you ask for guidance?

f(x)=a x is all obtained by 1 x scaling, and if you use it to discuss addition and subtraction, it obviously won't work!

The x in x2 belongs to [-1 4,1], which can be known as x2[1 16

1], that is, the x in y=f(x) belongs to [1 16

1], I didn't know this question when I was in high school at the beginning, because I always regarded X as the same, in fact, the key is that X in this is not the same meaning, and you can pay more attention to this point in the future. Functions are more difficult, but if you can learn them well, then you can do it.

There are two ways to think of it: (Note: x 3 represents the cubic of x) First, definition.

Let x10f(x1)-f(x2)=-x1 3+1-(-x2 3+1)=x2 3-x1 3

x2-x1) (x2 2+x1x2+x1 2) is held by the fundamental inequality a 2 + b 2 2ab (if and only if a=b is equal sign).

It can be inferred |a|^2+|b|^2≥2|ab|So a 2+b 2 2|ab|Here due to x1≠x2

Therefore x2 2+x1 2 2|x1x2|So x2 2+x1x2+x1 2 |x1x2|+(x1x2| +x1x2)＞0

So f(x1)-f(x2)=(x2-x1)(x2 2+x1x2+x1 2) 0

f(x) is a subtractive function.

2. Use the function image change.

This function is derived from the x3 image transformation, first about the y-axis fold becomes -x 3 and then translates one unit upwards to -x 3+1

x 3 can be seen to be an increase function, which becomes a subtraction function after folding, and then panning does not change the increase or decrease, so it finally becomes a subtraction function.

y=-x^3+1

It is possible to find a derivative.

y'=-3x 2<0 x ry decrement.

I used to be, but now I forgot.

Find the invert y=-x2, and the derivative is always less than zero on r, so it is single subtracted on r.

Since f(x+1)=-f(x), then f(x+2)=f[(x+1)+1]=-f(x+1)=-f(x)]=f(x), so f(x) is a periodic function, and the period is 2

When 1<=x<=2, -1<=(x-2)<=0 so f(x)=f[(x-2)+2]=f(x-2)=(x-2) 3-2(x-2)-1=(x-2) 3-2x+3

Functions, mainly transformations, commutations of the way of thinking are important.

The periodic function, mainly the definition, deformation, and the deformation of the first line, for example: f(x+2)=-1 f(x).

then, f(x+4)=. f(x)

。As an exercise, I believe you can do it.

This question doesn't need to be simplified, you can analyze it and come up with the answer.

The first thing you can determine is 0<2<, so in the interval (0, ), sinx>0.

And according to the curve of change of 2 to the power of x, in the interval (0,2), (2 to the power of x) > 1, then *<0 in (*)sinx. (* represents what you are in parentheses).

This allows it to be concluded that in the range (0,2), f(x) <0

In the same way, it can be analyzed that in the range (-2,0), f(x)<0.

So choose A

When x>0, there is -x<0

Again: when x 0, f(x)=x · 2 x

So there is: f(-x)=-x*2 (-x).

The odd function gets: f(x)=-f(-x).

So, when x>0, f(x)=-f(-x)=-[-x*2 (-x)]=x*2 (-x).

Let x 0, then -x 0, f(-x) (x)2（-x）

Another odd function. So f(x) -f(-x).

This is a fixed algorithm, which is to find which segment and which segment, and then convert it to a comprehensible interval.

Let -x<0, then x>0

f(-x)=(-x) ·2 (-x) The function y=f(x) is the odd function f(-x)=f(x) defined on r

f（x）=x · 2^(-x)

When x>0, there is -x<0

Again: when x 0, f(x)=x · 2 x

So there is: f(-x)=-x*2 (-x).

From the property of the odd function: f(x)=-f(-x), we get:

When x>0, f(x)=-f(-x).

[-x*2^(-x)]

x*2^(-x)

So f(x)=x*2 (-x) at x>0

f(x) is an odd function then f(-2)=-f(2).

f(x) is the additive function on the defined domain, so f(m 2-m) + f(-2) = f(m 2-m)-f(2)<0

i.e. m2-m<2 so-1

Since f(m 2-m) + f(-2) < 0, it is required that f(m 2-m)-f(2)<0 [f(x) is an odd function].

And because f(x) is an additive function on the defined domain, that is, m 2-m-2<0 is required, and the solution is -1

Because the function is odd.

So f(x) -f(-x).

So f(-2) = -f(2).

then substitute f(m2-m)-f(2)<0

f(m2-m) is known to be an increasing function.

m^2-m<2

Solution (m+1) (m-2) <0

1

f(m2-m)<-f(-2)=f(2) because it is an odd function and is an increasing function.

So m 2 - m < 2

Just find it.

First of all, when encountering such a problem, it is necessary to explore some hidden features of the equation by changing the form of x, so that x=1-t is brought into the original equation by bringing x into the original equation to obtain f(1-t)+2f(t)=1-t, because t is also an unknown number, so t and x are equivalent and can be interchangeably obtained.

Let's replace t with x to get it.

f(x)+2f(1-x)=x and f(1-x)+2f(x)=1-x subtract f(1-x) from the equation 2 to get f(x)=(2-3x) 3

Probably yes.

Let f(x)=kx+b

It is easy to know that f[f(x)]=f[kx+b]=k(kx+b)+b=k 2*x+kb+b=4x-1

then k 2 = 4 kb + b = -1

The solution gives k=2, b=-1, 3 or k=-2, b=1, so f(x)=2x-1 3 or f(x)=-2x+1

f(x)=kx+b

f[f(x)]=

k(kx+b)+b

k²x+b(k+1)

4x-1 so k = 4

k(b+1)=-1

k = 2 substituting k (b + 1) = -1 to find b

So f(x)=-2x-1 2 or f(x)=2x-3 2