Please help me solve a high school function problem for more details

The sufficient condition is that if the equation for x y=(ax 2+bx+c) (x 2+mx+n) has a real solution, then y is within the range of the function f(x).

The necessary condition is that if y is in the range of the function f(x), then the equation for x y=(ax 2+bx+c) (x 2+mx+n) must have a real solution.

The equation for x y=(ax 2+bx+c) (x 2+mx+n) has a real solution means that the equation for x (ax 2+bx+c) (x 2+mx+n)-y=0 (where y is known and only x is unknown) has a real solution.

That is, if the equation for x y=(ax 2+bx+c) (x 2+mx+n) has a real solution, then that y is in the range.

Sufficient means sufficient necessary means that the conclusion is valid (sufficient) as long as the condition is satisfied, and the conclusion must be satisfied (necessary).

This thing... What do you mean by your title?

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I'm not very familiar with excel formulas, please help me solve what the following formula means, thank you sumif function ......Does this explain it? Funny questioner, why did you cancel that question just now? Harm.

Very simple. 1. Because x is its real root, the corresponding y also has a value (the value range of y is the set of values of y obtained by bringing all the real numbers x into it); 2. If the value of y is in the value range, the value of x must be a real number, otherwise if it is a complex root, x is not in the real number field. (For real functions, the definition domain is in the field of real numbers).

1. All use the exchange method.

Let log2 x=t

f(x)=2(log2 x)2+2a log2 1/x +b =2t^2-2at+b

x=1 2.

log2 x=t=-1

2t 2-2at+b is a quadratic function.

The minimum value is produced on the axis of symmetry.

So a 2=-1

a=-2 is substituted into -1 to get the solution.

b=-6f(x)= =2t^2+4t-6>0

Solve t<-3 t>1

So log2 x=t>1 log2 x=t<-3x>2 0

Because f(x) is an odd function.

1. C=02, C is not equal to 0, then F(0)=0, so 2 C=0 has no solution. In summary, c = 0

From the meaning of the question, it is found that f(1 2)=2, and it is an odd function, then f(-1 loses the argument 2)=-2

Equation:。。

The solution yields a=1, b=2, c=0

|x|/(x+2)=k²x

1.x=0 and k belong to r

x=1 k 2 - 2 >0 , and the root number of -2 points is 23x<0, x=-1 k2 - 2 <0 k is not equal to 0

4.When there are 3 real roots, k is the intersection of the above 3 terms, and the 4th root is not found. . .

Odd function, f(0)=0, finds a=-1

So f(x)=log10((x+1)(1-x))<0=log10(1).

So 0<(x+1) (1-x)<1

Solution-11

First of all, f(0) = 0

Note that x≠1 shows that a = a 1

Order 0 2 (1 a x) a 1 1

So x 0

With the odd function, we know that f(0)=0, and find a=-1

So f(x)=log10 (2 (1-x)} a) is equal to f(x)=log10((x+1) (1-x))<0=log10(1).

So 0<(x+1) (1-x)<1

Solution-1

'(x)=2xe^(x-1)+(x^2)e^(x-1)+3ax^2+2bx

f'(-2)=0,f'(1)=0

Find a=-1 3, b=-1

x)=2xe^(x-1)+(x^2)e^(x-1)-x^2-2x=x(x+2)(e^(x-1)-1)

List. x(-infinity,-2)(-2,0)(0,1)(1,+infinity) book front.

f'(x) -

You can know the monotony of Bi Collapse, right? . .

Question 1: Derivation.

Question 2 Derivation first, analysis can obtain a>0

The following is a discussion of the derivative function.

First, the discriminant of the derivative function is less than or equal to 0

Second, if the axis of symmetry is less than 0, and the value of the function is greater than or equal to 0 at x=0, the third question is to write the expression h(x) first.

a<0 we find that h(0)>0 is constant, h(-1)>0 is constant and h(0) so we can get the position of the coordinate axis.

The positions of the axes (1) and (-1,0) are discussed below, and then the right intersection (the intersection point with the axes) is calculated

1.Substituting a=1 to find the derivative = 0, solve 3x 2 + 2x-1 = 0, x = -1 or 1 3

At this point f(x) = 1 or 15 27

Question 1 Associative Elementary Functions For example, defining the domain r, the range -1 to 3, is obviously constructable.

y=2sinx+1

Question 2 If you don't have a formal word, you can actually draw it

The formal practice is:

Let x be less than 0, then -x is greater than 0, bring -x as a whole into the f(x) analytic formula, and then use the odd function to find the analytic expression when x is less than 0.

Question 3 Outer centric, Pa=PB means that on the perpendicular bisector of AB, Pa=Pc is the same, and the intersection of these perpendicular bisectors is the outer centrocenter of the triangle.

In the fifth question, the latter subtracts the former to obtain the square of (x1-x2), which is always greater than or equal to 0, and the latter is greater than or equal to the former.

Fourth, I don't know what you ask, the set a represents the definition domain of the function x2-1, which is r, and then you can take a look.

I'm tired to death, and I expect me to be

On the interval [1,4], f(x)=x+4 x+af'(x)=1-4 x stationary x=2 left-right+, which is the minimum value of the minimum point f(x)=4+a

The maximum value of f(x) = max[f(1),f(4)] = 5 + a a = 0