Question

11．已知函數(shù)f（x）=$\left\{\begin{array}{l}{{2}^{-x}-1，x≤0}\\{\sqrt{x}，x＞0}\end{array}\right.$若f[f（x₀）]=1，則x₀=-1或1．

Answer 1

分析 當(dāng)x₀≤0時(shí)，$f（{x}_{0}）={2}^{-{x}_{0}}-1$，由f（x₀）=${2}^{-{x}_{0}}-1＜0$，得f[f（x₀）]=f（${2}^{-{x}_{0}}$-1）=${2}^{-{2}^{-{x}_{0}}-1}-1=1$，無(wú)解，由$f（{x}_{0}）={2}^{-{x}_{0}}-1$＞0，解得x₀=-1；當(dāng)x₀＞0時(shí)，f（x₀）=$\sqrt{{x}_{0}}$＞0，由f（f（x₀））=f（$\sqrt{{x}_{0}}$）=$\sqrt{\sqrt{{x}_{0}}}$=1，解得x₀=1．

解答 解：∵函數(shù)f（x）=$\left\{\begin{array}{l}{{2}^{-x}-1，x≤0}\\{\sqrt{x}，x＞0}\end{array}\right.$，f[f（x₀）]=1，
∴當(dāng)x₀≤0時(shí)，$f（{x}_{0}）={2}^{-{x}_{0}}-1$，
當(dāng)f（x₀）=${2}^{-{x}_{0}}-1＜0$時(shí)，f[f（x₀）]=f（${2}^{-{x}_{0}}$-1）=${2}^{-{2}^{-{x}_{0}}-1}-1=1$，無(wú)解，
當(dāng)$f（{x}_{0}）={2}^{-{x}_{0}}-1$＞0時(shí)，$\sqrt{{2}^{-{x}_{0}}-1}$=1，解得x₀=-1，成立；
當(dāng)x₀＞0時(shí)，f（x₀）=$\sqrt{{x}_{0}}$＞0，∴f（f（x₀））=f（$\sqrt{{x}_{0}}$）=$\sqrt{\sqrt{{x}_{0}}}$=1，解得x₀=1，成立．
綜上，x₀的值為-1或1．
故答案為：-1或1．

點(diǎn)評(píng) 本題考查實(shí)數(shù)值的求法，是基礎(chǔ)題，解題時(shí)要認(rèn)真審題，注意函數(shù)性質(zhì)的合理運(yùn)用．