How do I prove that f(x)=x2−4x+3 is injective at (−∞,2) and (2,∞)

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eninsala06 · Accepted Answer

Step 1  If  f(x)=f(y)  then  x2−4x=y2−4y.  Hence  x2−y2−4(x−y)=0,  which means  (x−y)(x+y−4)=0.  This is true if and only if x=y or x+y−4=0.  Now if x,y&amp;gt;2, then x+y&amp;gt;4, so the only possibility is x=y. Same for x,y&amp;lt;2 (try).

vicki331g8 · Accepted Answer

Step 1  On the on the interval (2,+∞), we have  f(x)=f(y)⇒(x+y)(x−y)=4(x−y),  suppose that x≠y, then we have  x+y=4, but we know that both x,y&amp;gt;2, a contradiction. Similar for the interval (−∞,2).

alenahelenash · Accepted Answer

Step 1This real valued function is obiously continuous and differentiable at every point in R.df(x)dx=2x−4.Sof′(x)&amp;gt;0,x∈(2,+∞)andf′(x)&amp;lt;0,x∈(−∞,2).So since f is properly monotonous and coninouous in those two specific intervals, is also injective, i.e. if x≠y then f(x)≠f(y) .

How do I prove that f(x)=x2−4x+3 is injective at (−∞,2) and (2,∞)

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