1) d
let's solve the equation f(x)=0
x2−5x+1=0x^2-5x+1=0x2−5x+1=0
x=5±212x=\frac{5\pm\sqrt{21}}{2}x=25±21
This equation has two roots, therefore this function is not one-to-one
2) d
f(x+Δx)−f(x)=(x+Δx)2−4(x+Δx)+7−x2+4x−7=x2+2xΔx+Δ2−4(x+Δx)−x2+4x=f(x+\Delta x)-f(x)=(x+\Delta x)^2-4(x+\Delta x)+7-x^2+4x-7=x^2+2x\Delta x+\Delta^2-4(x+\Delta x)-x^2+4x=f(x+Δx)−f(x)=(x+Δx)2−4(x+Δx)+7−x2+4x−7=x2+2xΔx+Δ2−4(x+Δx)−x2+4x=
=2xΔx+Δ2x−4Δx=2x\Delta x+\Delta^2x-4\Delta x=2xΔx+Δ2x−4Δx
\frac{}{} f(x+Δx)−f(x)Δx=2xΔx+Δ2x−4ΔxΔx=2x+Δx−4\frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{2x\Delta x+\Delta^2x-4\Delta x}{\Delta x}=2x+\Delta x-4Δxf(x+Δx)−f(x)=Δx2xΔx+Δ2x−4Δx=2x+Δx−4
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