QuestionFind values of $x$ for which $f(g(x)) = g(f(x))$ where $f(x)=\sqrt{x^{2}-1}$ and $g(x)=\sqrt{x^{2}+1}$ .

Studdy Solution

STEP 1

Assumptions1. The function $f(x)=\sqrt{x^{}-1}$ is defined for $x \geq1$ or $x \leq -1$ . The function $g(x)=\sqrt{x^{}+1}$ is defined for all real numbers $x$
3. We are looking for values of $x$ that make $f(g(x))$ and $g(f(x))$ commutative, i.e., $f(g(x)) = g(f(x))$

STEP 2

First, let's find $f(g(x))$ . We substitute $g(x)$ into $f(x)$ .
$f(g(x)) = f(\sqrt{x^{2}+1})$

STEP 3

Now, substitute $\sqrt{x^{2}+1}$ into $f(x)$ .
$f(g(x)) = \sqrt{(\sqrt{x^{2}+1})^{2}-1}$

STEP 4

implify the expression.
$f(g(x)) = \sqrt{x^{2}+1-1}$

STEP 5

Further simplify the expression.
$f(g(x)) = \sqrt{x^{2}}$

STEP 6

The square root of a square is the absolute value of the number, so we have $f(g(x)) = |x|$

STEP 7

Now, let's find $g(f(x))$ . We substitute $f(x)$ into $g(x)$ .
$g(f(x)) = g(\sqrt{x^{2}-1})$

STEP 8

Now, substitute $\sqrt{x^{2}-1}$ into $g(x)$ .
$g(f(x)) = \sqrt{(\sqrt{x^{2}-1})^{2}+1}$

STEP 9

implify the expression.
$g(f(x)) = \sqrt{x^{2}-+}$

STEP 10

Further simplify the expression.
$g(f(x)) = \sqrt{x^{2}}$

STEP 11

Again, the square root of a square is the absolute value of the number, so we have $g(f(x)) = |x|$

STEP 12

Now, we equate $f(g(x))$ and $g(f(x))$ to find the values of $x$ that make them commutative.
$f(g(x)) = g(f(x))$

STEP 13

Substitute the expressions we found for $f(g(x))$ and $g(f(x))$ .
$|x| = |x|$

STEP 14

This equation holds true for all real numbers $x$ . However, we must consider the domain of the original functions $f(x)$ and $g(x)$ . Since $f(x)$ is only defined for $x \geq$ or $x \leq -$ , these are the only values of $x$ that make $f(g(x))$ and $g(f(x))$ commutative.
So, the solution is $x \geq$ or $x \leq -$ .

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