Senior 1 math functions. Thank you

a=1f( x)=1--1/x) how to two minus signs.

This is mainly about making the intrinsic function a variable t.

Write the external function as an expression of t, and then convert t to an expression of x.

The rest is simplified.

These two questions should not be the same question, and I did so as such.

g[f(x)]=1/4(4x^2+4ax+a^2+3)=x^2+ax+(a^2+3)/4

By simplifying, we can get the equation like this: x+1=ax+(a 2+3) 4 Using the observation method, the two sides of the equation can be compared, a=1

f[f(x)]=1-1 (1-1 x) to obtain f[f(x)]=1 (1-x).

g[f(x)]=(（2x+a)^2+3)/4=(4x^2+4xa+a^2+3)/4=x^2+xa+(a^2+3)/4;

and g[f(x)]=xsquared+x+1;

So x 2+xa+(a 2+3) 4=x 2+x+1So a=1.

f[f(x)]=1-1 (1-1 x)=1 (1-x)Okay, that's it.

There are 2 problems: First, the first g[f(x)]=x2+x+12x+a)2+3=4*(x2+x+1).

4x2+4ax+a2+3=4x2+4x+4 is equal or so, so a=1

The second f[f(x)]=(-1) (x-1) If you have a question, you can continue to ask.

Solution: f(x)=ax +(3+a)x+3,a≠0

The function image is a parabola, the axis of symmetry is x=-(3+a) (2a), and the coordinates of the vertices are (-(3+a) (2a),3-(3+a) 4a)).

f(x)=(x+1)(ax+3)

The two intersections of the function with the x-axis are x=-1 and x=-3 a, respectively

Category Discussion: 1) a 0

a) At 0 a 3, f(x) increases on [-1,4] with a single-limb plexus feast, f(x)max=f(4)=16a +4a+15=4, and obtains, 16a +4a+11=0, no solution.

b) A 3, the axis of symmetry of f(x) is in the interval [-1,0], combined with the symmetry of the function.

f(x)max=f(4)=16a +4a+15=4, no solution.

2) a 0a) when -(3+a) (2a) 4, i.e., a -1 3, the function obtains the maximum value f(x)max=3-(3+a) 4a)=4 at the vertice, and obtains the calendar silver a +10a+9=0, and the solution is a=-1 or -9

b) When -(3+a) (2a) 4, i.e., 0 a -1 3, the function increases monotonically on [-1,4] between Zheng Ran, f(x)max=f(4)=16a +4a+15=4, and there is no solution.

In summary, the question can be satisfied when a=-1 or -9.

If you have any questions, please ask for this question, thank you.

f(x)=a[x+(3+a)/2a]^2+3-(3+a)^2/4a

1.When a<0, there are three cases.

1).For Li Juna, the axis of gravity is between [-1,4], f(x)max=3-(3+a) 2 4a=4, the solution is a=-9 or a=-1, and at this time, -1<=-3+a answer hole 2a) <=4, and a<=-1, then a=-9 and a=-1 satisfy the topic;

2).When the axis of symmetry is to the left of [-1,4], i.e., -(3+a2a)<=1, then f(x)max=f(-1)=4, a has no solution.

3).When the axis of symmetry is on the right side of [-1,4], that is, -(3+a2a)>=4, then f(x)max=f(4)=4, the solution a=-11 20 satisfies the problem;

2.When a>0, there are two scenarios.

1).The axis of symmetry x=-(3+a 2a)<=1+4) 2, where f(x)max=f(4)=4, a has no solution.

2).Symmetry which axis x=-(3+a 2a)>=1+4) 2, where f(x)max=f(-1)= no solution.

In summary, there is a, and when a=-9 or a=-1 or a=-11 20 fits the topic.

When x<=0, f(x)=1, and when x>0, f(x)=x 2+1 is monotonic.

The inequality f(1-x 2)>f(2x) yields:

1-x 2>2x and 1-x 2>0

The former gets x 2+2x-1<0, that is, the value range of -1-2 is -1

When x is less than zero, -x is greater than zero, so if you want to substitute -x into the expression of f(x), the first approach is wrong, and it is right to replace the x in x squared + x with -x.

Because you only know the case of x>0, and you don't have an analytic formula for the case of x<0. So when x<0, you don't have anything to bring in. That's why Law 1 isn't right.

Therefore, when x<0, the odd function f(x)=-f(-x)Since x<0, -x>0, then f(-x)=(-x) 2+(-x)=x 2-x

So at this point f(x)=-x 2+x

The first is wrong when it brings in negative x, and it ends up being wrong when it leads to finding the opposite number.

Simplify the following equation with a cubic formula.