Let f:R−2−1→R and g:R−2−5→R be defined as f(x)=2x+12x+3 and g(x)=2x+5∣x∣+1. Then, the domain of the function fog is :
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Official solution
f(x)=2x+12x+3;x=−21g(x)=2x+5∣x∣+1,x=−25
Domain of f(g(x))f(g(x))=2g(x)+12g(x)+3x=−25 and 2x+5∣x∣+1=−21x∈R−−25 and x∈R∴ Domain will be R−−25