Câu 3:
a: ĐKXĐ: x-1<>0
=>x<>1
=>Đúng
b: \(f\left(x\right)=\frac{x^2+4x-1}{x-1}\)
=>\(f^{\prime}\left(x\right)=\frac{\left(x^2+4x-1\right)^{\prime}\cdot\left(x-1\right)-\left(x-1\right)^{\prime}\left(x^2+4x-1\right)}{\left(x-1\right)^2}\)
\(=\frac{\left(2x+4\right)\left(x-1\right)-\left(x^2+4x-1\right)}{\left(x-1\right)^2}\)
\(=\frac{2x^2-2x+4x-4-x^2-4x+1}{\left(x-1\right)^2}=\frac{x^2-2x-3}{\left(x-1\right)^2}\)
=>Đúng
c: Đặt f'(x)=0
=>\(\frac{x^2-2x-3}{\left(x-1\right)^2}=0\)
=>\(x^2-2x-3=0\)
=>(x-3)(x+1)=0
=>x=3(nhận) hoặc x=-1(loại)
\(f\left(3\right)=\frac{3^2+4\cdot3-1}{3-1}=\frac{9+12-1}{2}=\frac{12+8}{2}=\frac{20}{2}=10\)
\(f\left(2\right)=\frac{2^2+4\cdot2-1}{2-1}=\frac{4+8-1}{1}=3+8=11\)
=>GTNN là 10
=>Sai
d: \(f\left(x\right)=\frac{x^2+4x-1}{x-1}\)
\(=\frac{x^2-x+5x-5+4}{x-1}=x+5+\frac{4}{x-1}\)
=>Tiệm cận xiên là y=x+5
=>Đúng
