f(x)= ax^( 2) + bx +c is given a ≠ b ≠ c , a , b , c ∈ R a≠0 and : f(ax + b )=f (bx + c) find : (1/2) (f(b) − f(a ))=?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 174096 by mnjuly1970 last updated on 24/Jul/22

$f (x) = {a x}^{2} + b x + c i s g i v e n$ $a \neq b \neq c, a, b, c \in R$ $a \neq 0 a n d :$ $f (a x + b) = f (b x + c)$ $f i n d : \frac{1}{2} (f (b) - f (a)) = ?$
	Answered by mahdipoor last updated on 24/Jul/22

	$i f f (x) = a x x + b x + c a n d$ $M a x / m i n f = f (\frac{- b}{2 a})$ $f (m) = f (n) \Leftrightarrow \frac{m + n}{2} = \frac{- b}{2 a}$ $\Rightarrow\Rightarrow \frac{(a x + b) + (b x + c)}{2} = \frac{- b}{2 a} \Rightarrow$ $a + b = 0 a n d b + c = \frac{- b}{a} \Rightarrow$ $a = k b = - k c = 1 + k \Rightarrow$ $f (x) = k x x - k x + 1 + k$ $\frac{1}{2} ([k {(- k)}^{2} - k (- k) + 1 + k] -$ $[k (k^{2}) - k (k) + 1 + k]) = k^{2} = a^{2}$
	Commented by mnjuly1970 last updated on 24/Jul/22

	$t h a n k y o u m a s t e r . . f a g h a t = a^{2}$ $t y p o . . .$
	Commented by mahdipoor last updated on 24/Jul/22

	$t n x, i e d i t e d$
	Answered by cortano1 last updated on 25/Jul/22
	$f(ax+b)=a(ax+b)^2 +b(ax+b)+c f(bx+c)=a(bx+c)^2 +b(bx+c)+c { ((f(ax+b)=a^3 x^2 +2a^2 bx+ab^2 +abx+b^2 +c)),((f(bx+c)=ab^2 x^2 +2abcx+ac^2 +b^2 x+bc+c)) :} { ((a^2 =b^2 ⇒ { ((a=b(rejected))),((a=−b)) :})),((2a^2 b+ab=b^2 +2abc)),((ab^2 +b^2 +c=ac^2 +bc+c)) :} for a=−b ⇒ { ((−2a^3 −a^2 =a^2 −2a^2 c)),((a^3 +a^2 +c=ac−ac+c)) :} ⇒a^2 (a+1)=0⇒a=−1 , b=1 , ⇒2−1=1−2c ; c=0 ∴ f(x)=−x^2 +x { ((f(b)=f(1)=−1+1=0)),((f(a)=f(−1)=−1−1=−2)) :} ⇒ (1/2)[ f(b)−f(a) ]= (1/2)(0−(−2))=1$
	$f (a x + b) = a {(a x + b)}^{2} + b (a x + b) + c$ $f (b x + c) = a {(b x + c)}^{2} + b (b x + c) + c$ ${\begin{cases} f (a x + b) = a^{3} x^{2} + 2 a^{2} b x + {a b}^{2} + a b x + b^{2} + c \\ f (b x + c) = {a b}^{2} x^{2} + 2 a b c x + {a c}^{2} + b^{2} x + b c + c \end{cases}$ ${\begin{cases} a^{2} = b^{2} \Rightarrow {\begin{cases} a = b (r e j e c t e d) \\ a = - b \end{cases} \\ 2 a^{2} b + a b = b^{2} + 2 a b c \\ {a b}^{2} + b^{2} + c = {a c}^{2} + b c + c \end{cases}$ $f o r a = - b \Rightarrow {\begin{cases} - 2 a^{3} - a^{2} = a^{2} - 2 a^{2} c \\ a^{3} + a^{2} + c = a c - a c + c \end{cases}$ $\Rightarrow a^{2} (a + 1) = 0 \Rightarrow a = - 1, b = 1,$ $\Rightarrow 2 - 1 = 1 - 2 c; c = 0$ $∴ f (x) = - x^{2} + x$ ${\begin{cases} f (b) = f (1) = - 1 + 1 = 0 \\ f (a) = f (- 1) = - 1 - 1 = - 2 \end{cases}$ $\Rightarrow \frac{1}{2} [f (b) - f (a)] = \frac{1}{2} (0 - (- 2)) = 1$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum