Senior 3 Function Problems, Senior 3 Functional Math Problems

f(1+x)=f(1-x) indicates that the parabola is symmetrical with x=1, then -b 2a=1, b=-2a

Let f(x)=ax 2-2ax+ac=a(x 2-2x+c) (Note: The constant term is ac for the convenience of the following calculations).

Let x1 and x2 be two, then.

x1+x2=2,x1*x2=c

The sum of two cubes of f(x)=0 is 17

x1^3+x2^3

x1+x2)[(x1+x2)^2-3x1x2]2*(4-3c)

c=-3/2

The maximum value of f(x) is 15, which means that a<0 and f(1)=1515=a(1-2+c).

Substituting c=-3 2 into the :

a=-6 f(x)=a(x^2-2x+c)=-6(x^2-2x-3/2)=-6x^2+12x+9

Analytical formula of f(x): f(x)=-6x 2+12x+9

The three conditions given by the question.

1）f(1+x)=f(1-x)

f(1+1-x)=f(x)

f(x)=f(2-x)

It is said that the axis of symmetry of the function is 1

2) f(x) maximum value of 15

There is a maximum value indicating that the quadratic term of the function is less than 0

and f(1)=15

3) f(x)=0, the sum of two cubes is 17

Let f(x)=ax 2+bx+c(a<0).

f(1)=a+b+c=15

b/2a=1 => b=-2a>0

x1+x2=-b/a,x1x2=c/a

x1 3+x2 3=(x1+x2)(x1 2-x1x2+x2 2)(x1+x2)[(x1+x2) 2-3x1x2]-b a*[b 2 a 2-3c a]=17 Simultaneous solution of three equations.

a=-15/4

b=15/2

c=45/4

So the analytic formula of the function is.

f(x)=-15/4x^2+15/2x+45/4

(1) Set a==

It means that when f(x)=x, that is, ax 2+(b-1)x+c=0, the two x1 and x2 are 1 and 2, respectively

So from Vedder's theorem, we get x1+x2=-(b-1) a=3x1*x2=c a=2

Then from f(0)=c=2, we get a=1, b=-2, c=2, so f(x)=x 2-2x+2=(x-1) 2+1 from the f(x) image can be known:

m=f(x)min=f(1)=1

m=f(x)max=f(-2)=10

2) Set a==

It means that when f(x)=x, that is, ax 2+(b-1)x+c=0, there is only one root x=2

So we get =(b-1) 2-4ac=0

and x=-b 2a=2

Substituting it yields b=-4a, c=[(b-1) 2] 4a=4a+1 (4a)+2

Then f(x)=ax 2-4ax+4a+1 (4a)+2 because a 1, the fundamental inequality can be used.

f(x) ax 2-4ax+2[ 4a*1 (4a)]+2=ax 2-4ax+4

The axis of symmetry x=-b 2a=-(-4a) 2a=2 can be seen from the image of f(x).

m=f(x)min=f(2)=-4a+4

m=f(x)max=f(-2)=12a+4, then g(a)=m+m=8a+8 (a 1) from the image of g(a): g(a) increases monotonically on [1,+, so g(a)min=g(1)=8*1+8=16

Solution: From the problem setting, it can be seen that on the interval [1,2], there is always :

x|x-a|+2x-3＜2x-2

Constant: x|x-a|＜1

x-a|＜1/x

1/x＜x-a＜1/x

When 1 x 2, there is always x-(1 x) a x+(1 x) constructor g(x)=x+(1 x), h(x)=x-(1 x) 1≤x≤2

It is easy to know, there is always 2 g(x) 5 20 h(x) 3 2 Given by the question: 3 2 a 2

x|x-a|+2x-3<2x-2 is constant, equivalent to x|x-a|<1

1 xx-1 x because x-1 x increments over x [1,2], the maximum value is x-1 x=2-1 2=3 2

x+1 x increments over x [1,2], and the minimum value is 2 x+1 x=1+1 1=2

So 3 2

x|x-a|+2x-3<2x-2, then |x-a|<1 x,-1 x let g(x)=x-1 x, h(x)=x+1 x, when 1<=x<=2, the maximum value of g(x) is 3 2, and the minimum value of h(x) is 2.

Therefore, the range of values of the real number a is [3, 2, 2].

f'(x)=e+(1 x)+4x+m>0 is held at (0,+ (note that the original question is ex not e x).

f"(x)=4-1/x^2

x=1 2 when the second-order derivative is 0, and the third-order derivative is f'''x)=2x (-3)>0, so the first-order derivative has a minimum value =e+4+m>0

So m>-4-e

p is a necessary condition for q.

m>=(1 x-4x-e)=-4-e makes use of mean inequalities.

1 x-4x = -1 x+4x) 1 x+4x > = 2 times 1 x multiplied by 4x = 2 times root 4 = 4

The above results are out.

Finding the derivative of f yields f'(x)=(x 2 - 2mx + 1) x 2, in order for f(x) to have two extremums, its derivative must have more than two zeros, i.e., f'(x)=0 has at least two roots. Since the defined field is x>0, let f'(x)=0, we get x 2 - 2mx + 1 = 0, which has two different solid roots, so (2m) 2-4*1*1 = 4m 2-4>0, so m>1 or m<-1;

From the relationship between the root and the coefficient, we can solve x1=m+ (m 2-1), x2=m- (m 2-1) (is the root number); x1+x2=2m，x1x2=1；

k = (f(x1)-f(x2))/(x1-x2) = 1+(1/x1x2)-2m*ln(x1/x2)/(x1-x2)

1 + 1 - m*ln( 2m^2 + 2m√( m^2-1) -1) ；

Let the above equation = 2-2m, then 1 + 1 - m*ln( 2m 2 + 2m ( m 2-1) -1) (m 2-1) = 2-2m

i.e. ln(2m 2+2m (m 2-1)-1) = 2 (m 2-1).

G(m) = ln(2m 2+2m (m 2-1)-1) -2 (m 2-1).

Finding a derivative of it yields g'(m) = 2 - 2m/√(m^2-1)

Ream'(m)=0, the equation has no solution. That is, when m>1, g'(m) <0, m<-1, g'(m) >0, g(m) increases in the interval less than -1 and decreases in the interval greater than 1.

When m=1 g(m)=0, since monotonically decreasing on (1,+) and increasing monotonically on (- 1), g(m)=0 has no other root than 1. And since m>1 or m<-1 is found earlier, g(m)=0 has no solution. Therefore there is no m such that k = 2-2m

First x>0, and f'(x)=(x 2-2mx+1) x 2=0 has two unequal positive roots, so 4m 2-4>0, and 2m>0&1>0 is solved to m>1; At this time, x1+x2=2m, x1x2=1; (One less braille in the original question), f(x1)-f(x2)=x1-x2-(1 x1-1 x2)-2m(lnx1 x2)=2(x1-x2)-4mlnx1, so k=2-4mlnx1 (2x1-m), if k=2-2m, then 2lnx1=2x1-m, where x1=m+-root number m 2-1, m=x1+1 x1) 2, move to obtain:; Let h(x)=, then h'(x)=3 2+1 2x 2-2 x=(3x 2-2x+1) x 2=0 has no solution, so h'(x)>0, h(x) is an increasing function, and x1>1 or x1<1, so h(x1), go down first, and then answer when you are free

The original question is: f(x)=(-3 x+a) (3 (x+1)+b)).If y=f(x) defines the domain as r, judge its monotonicity on r and prove it.

Solution: The domain is defined by f(x) as r gets:b 0 and f(x) is deformed to get :

f(x)=m (3 x+b 3)-1 3 where m=(3a+b) 9,b 0 f'(x)=(-m)(ln3)3 x (3 x+b 3) 2 where (ln3)3 x (3 x+b 3) 2>0 So when m=0 i.e. a=-b 3 f(x)=-1 3 is a constant function, not monotonic; When m>0 i.e. a>-b 3 f'(x)<0 and f(x) is a subtractive function over r; When m<0 is a3 then f'(x)>0 and f(x) is an increment function over r. The above methods are a simpler way to deal with this kind of problem in secondary school, I hope it will help you a little!

From the nature of the logarithmic function, we get that lg(), that is, all operations in () are greater than 0, that is, 1 1 x 0, and because x is the denominator, according to the definition of fractions, the denominator cannot be zero, that is, x 0, so the calculation is modified by the red pen in the figure above (I hope it helps you, thank you(

According to the title, 1-1 x 0, x ≠ 0, from 1 1 x, x 1 and x 0 can be solved, this is not difficult to solve.

Or draw a diagram to solve it, so that y1 = 1, y2 = 1 x, to y1 y2, that is, the area below y1 is satisfied, y1 is a straight line parallel to the x axis, y2 is two hyperbolas, and its definition domain is (negative infinity, 0) and (0, positive infinity), so y1 y2, that is, x 0 and x 1, hope to adopt.

Please click to adopt.

Select d 1-1 x 0, x 0 or 1, and the function y=lg(1-1x) defines the field:

From the question, a+c=1 2, c-1 16a 0, so 1 2-(a+1 16a) 0, and a+1 16a 1 2 (mean inequality) so a+1 16a=1 2 so a=c=1 4

g(x)=[x-(2m+1)] 2 4-m 2-m, when x=m+1 2, take the smallest, at this time -m 2-m=-5, so m=(-1 root number 21) 2

According to the image of the function y=k x, the image of the function is symmetrical with respect to the origin (k is constant). Similar to inverse scale images.

Translate this image two units to the left, and we get y=k (x+2), symmetry about (2,0).

Then translate the resulting image upwards by one unit to obtain y=1+k (x+2), symmetry about (2,1).

So y=1+k (x+2) is symmetrical with respect to (2,1), ie.

y=(x+2+k)/(x+2)

So k = 1 and a = 2

or y=(x+3) (x+a)=(x+a+3-a) (x+a)=1+(3-a) (x+a).

This function translates a|, from y=1 x to the left (a>0) or to the right (a<0).units, and then translate upwards by 1 unit, so the center of symmetry (0,0) also has to be translated in this way, and (2,1) means that the translation to the left is 2 units, and the upward translation is 1 unit, so a=2

The center of symmetry by y=(x+3) (x+a) is (-2,1).

That is, the two asymptotic lines of the hyperbola are x=-2 and y=1 (the standard equation can be obtained by rotation and translation).

By x+a=0, substitute x=-2 with: a=2

p：f'(x) = 2x-2m because f(x) is in [2,+.on monotonically increasing, so f'(2) >=0, i.e. m<=2

q: Because the inequality is always greater than 0, according to the knowledge of the quadratic function image, < 0, the image floats above the x-axis, and the function value is always greater than 0, and the solution is 1 m 3 from <0

That is, p:m<=2

q: 1=3, and the solution is m<=1

2.If p is false q true, then m > 2,1< m<3, the solution is 2