AskSia

Plus

8. 已知函数

f(x)

的定义域为

\mathrm{R}, f(x)>f(x-1)+f(x-2)

, 且当

x<3

时,

f(x)=x

,则下列...

Aug 14, 2024

Generated Graph

Solution by Steps

step 2

For

x = 0, 1, 2

, we have

f(0) = 0

f(1) = 1

, and

f(2) = 2

. We can calculate

f(3)

using the inequality: f(3) > f(2) + f(1) = 2 + 1 = 3. Thus,

f(3) \geq 4

step 3

Continuing this process, we find that f(4) > f(3) + f(2) \geq 4 + 2 = 6, so

f(4) \geq 7

step 4

By induction, we can show that f(n) > f(n-1) + f(n-2) leads to

f(n) \geq 2^{n-2}

for

n \geq 3

step 5

Therefore, for

n = 10

f(10) \geq 2^{10-2} = 2^8 = 256

, and for

n = 20

f(20) \geq 2^{20-2} = 2^{18} = 262144

. Thus, f(10) < 1000 and f(20) > 1000

Key Concept

Inequalities and recursive functions

Explanation

The function's growth is determined by its recursive definition, leading to exponential growth beyond certain points. This allows us to evaluate the function's values at specific points.