f(x)=(x+3)^2 My work: y=(x+3)(x+3) y=x^2+3x+3x+9 y=x^2+6x+9 x=y^2+6y+9 y^2+6y=x-9 I can't isolate y.
C Chocolate New member Joined Sep 2, 2005 Messages 30 Oct 31, 2005 #1 f(x)=(x+3)^2 My work: y=(x+3)(x+3) y=x^2+3x+3x+9 y=x^2+6x+9 x=y^2+6y+9 y^2+6y=x-9 I can't isolate y.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Nov 1, 2005 #2 Hello, Chocolate! Find the inverse function of: f(x) = (x + 3)2\displaystyle f(x) \,= \,(x\,+\,3)^2f(x)=(x+3)2 Click to expand... We have: . (x + 3)2 = y\displaystyle (x\,+\,3)^2 \,= \,y(x+3)2=y Then: . (y + 3)2 = x\displaystyle (y\,+\,3)^2 \,= \,x(y+3)2=x Take square roots: . y + 3 = ±x\displaystyle y\,+\,3 \,= \,\pm\sqrt{x}y+3=±x Hence: . y = −3 ± x\displaystyle y \:= \:-3\,\pm\,\sqrt{x}y=−3±x Therefore: . f−1(x) = −3 ± x\displaystyle f^{-1}(x) \:= \:-3\,\pm\,\sqrt{x}f−1(x)=−3±x . . (Note that the inverse is not a function.)
Hello, Chocolate! Find the inverse function of: f(x) = (x + 3)2\displaystyle f(x) \,= \,(x\,+\,3)^2f(x)=(x+3)2 Click to expand... We have: . (x + 3)2 = y\displaystyle (x\,+\,3)^2 \,= \,y(x+3)2=y Then: . (y + 3)2 = x\displaystyle (y\,+\,3)^2 \,= \,x(y+3)2=x Take square roots: . y + 3 = ±x\displaystyle y\,+\,3 \,= \,\pm\sqrt{x}y+3=±x Hence: . y = −3 ± x\displaystyle y \:= \:-3\,\pm\,\sqrt{x}y=−3±x Therefore: . f−1(x) = −3 ± x\displaystyle f^{-1}(x) \:= \:-3\,\pm\,\sqrt{x}f−1(x)=−3±x . . (Note that the inverse is not a function.)
