fog=6x^2 −2x+3 gof=12x^2 −14x+6 f(2)=?


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 165021 by mathlove last updated on 25/Jan/22

$f o g = 6 x^{2} - 2 x + 3$ $g o f = 12 x^{2} - 14 x + 6$ $f (2) = ?$
	Answered by mr W last updated on 25/Jan/22

	$f (x) = {a x}^{2} + b x + c$ $g (x) = p x + q$ $f (g (x)) = a {(p x + q)}^{2} + b (p x + q) + c$ $f (g (x)) = {a p}^{2} x^{2} + p (2 a q + b) x + {a q}^{2} + b q + c = 6 x^{2} - 2 x + 3$ ${a p}^{2} = 6$ $p (2 a q + b) = - 2$ ${a q}^{2} + b q + c = 3$ $g (f (x)) = p ({a x}^{2} + b x + c) + q$ $g (f (x)) = {p a x}^{2} + p b x + p c + q = 12 x^{2} - 14 x + 6$ $p a = 12$ $p b = - 14$ $p c + q = 6$ $12 p = 6 \Rightarrow p = \frac{1}{2} \Rightarrow a = 24$ $\frac{1}{2} b = - 14 \Rightarrow b = - 28$ $\frac{1}{2} (48 q - 28) = - 2 \Rightarrow q = \frac{1}{2}$ $24 \times \frac{1}{4} - 28 \times \frac{1}{2} + c = 3 \Rightarrow c = 11$ $\frac{1}{2} \times 11 + \frac{1}{2} = 6 ✓$ $f (x) = 24 x^{2} - 28 x + 11$ $g (x) = \frac{x}{2} + \frac{1}{2}$ $f (2) = 24 \times 2^{2} - 28 \times 2 + 11 = 51$
	Commented by aleks041103 last updated on 25/Jan/22

	$t h e r e i s o n e m o r e s o l u t i o n$
	Commented by mr W last updated on 25/Jan/22

	$y e s, y o u a r e r i g h t . t h a n k s!$ $i a l s o r e a l i s e d t h a t t h e r e a r e t w o$ $p o s s i b i l i t i e s .$
	Commented by Tawa11 last updated on 25/Jan/22

	$Weldone sirs$
	Answered by aleks041103 last updated on 25/Jan/22
	$Suplement to mr W answer suppose f(x)=ax+b g(x)=cx^2 +dx+e f○g(x)=a(cx^2 +dx+e)+b= =acx^2 +adx+(ae+b) g○f(x)=c(ax+b)^2 +d(ax+b)+e= =ca^2 x+2abcx+b^2 c+adx+bd+e= =a^2 cx^2 +(2abc+ad)x+(b^2 c+bd+e) ⇒ { ((ac=6)),((ad=−2)),((ae+b=3)),((a^2 c=12)),((2abc+ad=−14)),((b^2 c+bd+e=6)) :} (1) and (4)⇒a=((a^2 c)/(ac))=((12)/6)=2 ⇒ c=3 a=2⇒d=−1 ⇒ { ((2e+b=3)),((12b−2=−14⇒b=−1)),((3b^2 −b+e=6)) :} (1) and (2)⇒e=2 and 3.1+1+2=6✓ ⇒ { ((a=2)),((b=−1)),((c=3)),((d=−1)),((e=2)) :} ⇒f(x)=2x−1 and g(x)=3x^2 −x+2 f(2)=3$
	$S u p l e m e n t t o m r W a n s w e r$ $s u p p o s e$ $f (x) = a x + b$ $g (x) = {c x}^{2} + d x + e$ $f \circ g (x) = a ({c x}^{2} + d x + e) + b =$ $= {a c x}^{2} + a d x + (a e + b)$ $g \circ f (x) = c {(a x + b)}^{2} + d (a x + b) + e =$ $= {c a}^{2} x + 2 a b c x + b^{2} c + a d x + b d + e =$ $= a^{2} {c x}^{2} + (2 a b c + a d) x + (b^{2} c + b d + e)$ $\Rightarrow {\begin{cases} a c = 6 \\ a d = - 2 \\ a e + b = 3 \\ a^{2} c = 12 \\ 2 a b c + a d = - 14 \\ b^{2} c + b d + e = 6 \end{cases}$ $(1) a n d (4) \Rightarrow a = \frac{a^{2} c}{a c} = \frac{12}{6} = 2 \Rightarrow c = 3$ $a = 2 \Rightarrow d = - 1$ $\Rightarrow {\begin{cases} 2 e + b = 3 \\ 12 b - 2 = - 14 \Rightarrow b = - 1 \\ 3 b^{2} - b + e = 6 \end{cases}$ $(1) a n d (2) \Rightarrow e = 2$ $a n d 3.1 + 1 + 2 = 6 ✓$ $\Rightarrow {\begin{cases} a = 2 \\ b = - 1 \\ c = 3 \\ d = - 1 \\ e = 2 \end{cases}$ $\Rightarrow f (x) = 2 x - 1 a n d g (x) = 3 x^{2} - x + 2$ $f (2) = 3$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum