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Algebra Difficulty 2.9 Junior Find the answer

Given the function f(x)={2x,x<1x+3,x1f(x) = \begin{cases} 2^{x}, & x < 1 \\ -x+3, & x\geqslant 1 \end{cases}, find the value of f[f(0)]f[f(0)],

Pick one

Solution

Since the function f(x)={2x,x<1x+3,x1f(x) = \begin{cases} 2^{x}, & x < 1 \\ -x+3, & x\geqslant 1 \end{cases},

First, we find f(0)f(0). According to the function definition, when x<1x<1, f(x)=2xf(x) = 2^{x}. So, f(0)=20=1f(0) = 2^{0} = 1.

Next, we find f[f(0)]f[f(0)], which is equivalent to finding f(1)f(1). According to the function definition, when x1x\geqslant 1, f(x)=x+3f(x) = -x+3. So, f(1)=1+3=2f(1) = -1+3 = 2.

Therefore, the value of f[f(0)]f[f(0)] is 2\boxed{2}.

This problem tests our understanding of how to find the value of a function at a given point. It's important to carefully read the problem and make reasonable use of the properties of the function.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.