If $s (x) = x^{2} + 1$ $s(x)=x^2+1$ and $t (x) = x - 3$ $t(x)=x-3$ , does $s (t (x)) = t (s (x))$ $s(t(x))=t(s(x))$ ?

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If $s (x) = x^{2} + 1$ $s(x)=x^2+1$ and $t (x) = x - 3$ $t(x)=x-3$ , does $s (t (x)) = t (s (x))$ $s(t(x))=t(s(x))$ ?

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Answer 1 · 2017-03-01T00:30:01

No

Explanation:

Given
$XXX s (x) = x^{2} + 1$ $color(white)("XXX")color(red)(s(x))=color(red)(x^2+1)$
and
$XXX t (x) = x - 3$ $color(white)("XXX")color(blue)(t(x))=color(blue)(x-3)$

Then
$XXX s (t (x)) = {(t (x))}^{2} + 1$ $color(white)("XXX")s(color(blue)(t(x)))=(color(blue)(t(x)))^2+1$
$XXXXXXX = {(x - 3)}^{2} + 1$ $color(white)("XXXXXXX")=(color(blue)(x-3))^2+1$
$XXXXXXX = x^{2} - 6 x + 8$ $color(white)("XXXXXXX")=x^2-6x+8$
and
$XXX t (s (x)) = x^{2} + 1 - 3$ $color(white)("XXX")t(color(red)(s(x)))=color(red)(x^2+1)-3$
$XXXXXXX = x^{2} - 2$ $color(white)("XXXXXXX")=x^2-2$

Clearly $s (t (x)) \neq t (s (x))$ $s(t(x))!=t(s(x))$

Answer 2 · 2017-03-01T00:33:34

No.
$s (t (x)) = x^{2} - 6 x + 10$ $s(t(x)) = x^2-6x+10$ , while $t (s (x)) = x^{2} - 2$ $t(s(x)) = x^2-2$

Explanation:

$s (t (x)) = {(x - 3)}^{2} + 1$ $s(t(x)) =(x-3)^2+1$ (plug $t (x)$ $t(x)$ in the place of x in $s (x)$ $s(x)$
$s (t (x)) = x^{2} - 6 x + 9 + 1$ $s(t(x)) = x^2-6x+9+1$ (square $(x - 3)$ $(x-3)$ )
$s (t (x)) = x^{2} - 6 x + 10$ $s(t(x)) = x^2-6x+10$ (simplify)

$t (s (x)) = (x^{2} + 1) - 3$ $t(s(x)) = (x^2+1)-3$ (plug in $s (x)$ $s(x)$ in the place of x in $t (x)$ $t(x)$ )
$t (s (x)) = x^{2} - 2$ $t(s(x)) = x^2-2$ (simplify)

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If $s (x) = x^{2} + 1$ $s(x)=x^2+1$ and $t (x) = x - 3$ $t(x)=x-3$ , does $s (t (x)) = t (s (x))$ $s(t(x))=t(s(x))$ ?

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If s(x)=x2+1s(x)=x^2+1 and t(x)=x−3t(x)=x-3, does s(t(x))=t(s(x))s(t(x))=t(s(x))?

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If $s (x) = x^{2} + 1$ $s(x)=x^2+1$ and $t (x) = x - 3$ $t(x)=x-3$ , does $s (t (x)) = t (s (x))$ $s(t(x))=t(s(x))$ ?