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Question Number 186503 by mnjuly1970 last updated on 05/Feb/23
function of , f (x) = ax + ∣ x ∣ is one to one .find ” a ” .
functionof,f(x)=ax+∣x∣isonetoone.find″a″.
Commented by Frix last updated on 05/Feb/23
f(x)= { (((a−1)x; x<0)),(((a+1)x; x≥0)) :} f′(x)= { ((a−1; x<0)),((a+1; x≥0)) :} f(0)=0 We need (1) a−1<0∧a+1<0 or (2) a−1>0∧a+1>0 ⇒ a<−1∨a>1
f(x)={(a−1)x;x<0(a+1)x;x⩾0f′(x)={a−1;x<0a+1;x⩾0f(0)=0Weneed(1)a−1<0∧a+1<0or(2)a−1>0∧a+1>0⇒a<−1∨a>1

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