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Reflections on solving math problems in senior one.
1, answer: There are even and odd functions.

Analysis:

This is a formula because f(x+y)=f(x)-f(y), so x=y=0, f(0)=f(0)-f(0), so f(0)=0.

Let X=0, then f(y)=-f(y), so this is a parity function.

2. Answer: a= 1

Analysis:

F(x) is an even function, f( 1)=f(- 1)=0, and f( 1)=2 times (1-a)=0, so a= 1.

3. Answer: X belongs to (negative infinity, -3), (3, positive infinity),

Analysis:

Because f (-3) = 0 and f is odd function, so -f(3)=0, so f(3)=0,

When xf (x) < 0, the solution set is the first x < 0, f (x) > 0 or the second x > 0, f (x) < 0.

If it is the first type, the result from the image is (3, positive infinity).

In the second case, the result obtained from the image is (negative infinity, -3).

4. Answer: f (x) = cube of x-65438 +0.

Analysis:

Odd function has -f(x)=f(-x), x > 0, and f(x)=x cube+1.

At this time, -x < 0, let t =-x, t < 0, -t is greater than 0, and f(-t)= t+ 1 =-f(t) cube,

So f (t) = cube of t-65438 +0. When x is less than 0, just like t, replacing t with x results in the cubic (x) of f = x-1.