Let g(x)=x-3. Find a function f so that f(g(x))=$\displaystyle x^2$
Let f(x)=$\displaystyle x^2$. Find a function g so that f(g(x))=$\displaystyle x^2+8x+16$
Thanks!
We know that
$\displaystyle f(x - 3) = x^2$
Assume we have a function f(x). Then f(x - 3) is a translation of this function 3 units to the right. And we know this function is $\displaystyle x^2$. So if we translate the function f(x - 3) by 3 units to the left, we get f(x). So
$\displaystyle f(x) = (x + 3)^2$
-Dan
Hello, johett!
Let $\displaystyle g(x)=x-3$
Find a function $\displaystyle f(x)$ so that: .$\displaystyle f(g(x)) \:=\: x^2$
We want: .$\displaystyle f(x-3) \:=\:x^2$
. . $\displaystyle f(x)$ must transform $\displaystyle x-3$ into $\displaystyle x^2.$
This can be done by: .adding 3 and squaring.
Therefore: .$\displaystyle f(x) \:=\:(x+3)^2$
Let $\displaystyle f(x)\:=\:x^2$.
Find a function $\displaystyle g(x)$ so that: .$\displaystyle f(g(x)) \:=\: x^2+8x+16$
We want: .$\displaystyle f(g(x)) \:=\:(x+4)^2$
Therefore: .$\displaystyle g(x)\:=\:x+4$