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Question Number 208453 by hardmath last updated on 16/Jun/24

If   f(x) = (((2a + 1)∙x + 1)/(x − a))   and   f(x) = f^(−1) (x)  Find:   a^2  + 3 = ?

Iff(x)=(2a+1)x+1xaandf(x)=f1(x)Find:a2+3=?

Answered by efronzo1 last updated on 16/Jun/24

 f^(−1) (x)=((ax+1)/(x−(2a+1))) = (((2a+1)x+1)/(x−a))   (ax+1)(x−a)= (x−(2a+1))((2a+1)x+1)    ax^2 −(a^2 +1)x−a=(2a+1)x^2 +(1−(2a+1)^2 )x−(2a+1)    ⇒a=2a+1 ; a=−1    ⇒a^2 +3 = 4

f1(x)=ax+1x(2a+1)=(2a+1)x+1xa(ax+1)(xa)=(x(2a+1))((2a+1)x+1)ax2(a2+1)xa=(2a+1)x2+(1(2a+1)2)x(2a+1)a=2a+1;a=1a2+3=4

Answered by MM42 last updated on 16/Jun/24

f(f(0))=0⇒f(−(1/a))=0  ⇒(((2a+1)(−(1/a))+1)/(−(1/a)−a))=0  ⇒((2a+1)/a)=1⇒a=−1⇒a^2 +3=4 ✓

f(f(0))=0f(1a)=0(2a+1)(1a)+11aa=02a+1a=1a=1a2+3=4

Commented by hardmath last updated on 16/Jun/24

thankyou dear professor

thankyoudearprofessor

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