Find \(f \circ g\) and \(g \circ f\) for Given Absolute Value Functions
📺 Video Explanation
📝 Question
Let:
\[ f(x)=|x|+x,\qquad g(x)=|x|-x,\qquad x\in\mathbb{R} \]
Find:
- \((f\circ g)(x)\)
- \((g\circ f)(x)\)
Also find:
\[ (f\circ g)(-3),\quad (f\circ g)(5),\quad (g\circ f)(-2) \]
✅ Solution
🔹 Step 1: Simplify \(f(x)\) and \(g(x)\)
Using modulus:
For \(x\ge0\):
\[ |x|=x \]
So:
\[ f(x)=x+x=2x,\qquad g(x)=x-x=0 \]
For \(x<0\):
\[ |x|=-x \]
So:
\[ f(x)=-x+x=0,\qquad g(x)=-x-x=-2x \]
Thus:
\[ f(x)= \begin{cases} 2x,& x\ge0\\ 0,& x<0 \end{cases} \qquad g(x)= \begin{cases} 0,& x\ge0\\ -2x,& x<0 \end{cases} \]
🔹 Step 2: Find \((f\circ g)(x)\)
\[ (f\circ g)(x)=f(g(x)) \]
Since \(g(x)\ge0\) for all \(x\), use:
\[ f(t)=2t\quad (t\ge0) \]
So:
\[ (f\circ g)(x)=2g(x) \]
Now:
\[ g(x)=|x|-x \]
Therefore:
\[ \boxed{(f\circ g)(x)=2(|x|-x)} \]
🔹 Step 3: Find \((g\circ f)(x)\)
\[ (g\circ f)(x)=g(f(x)) \]
Since \(f(x)\ge0\) for all \(x\), use:
\[ g(t)=0\quad (t\ge0) \]
So:
\[ \boxed{(g\circ f)(x)=0} \]
🔹 Step 4: Evaluate values
(i) For \(x=-3\):
\[ (f\circ g)(-3)=2(|-3|-(-3))=2(3+3)=12 \]
\[ \boxed{(f\circ g)(-3)=12} \]
(ii) For \(x=5\):
\[ (f\circ g)(5)=2(|5|-5)=2(5-5)=0 \]
\[ \boxed{(f\circ g)(5)=0} \]
(iii) For \(x=-2\):
\[ (g\circ f)(-2)=0 \]
\[ \boxed{(g\circ f)(-2)=0} \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=2(|x|-x)} \]
\[ \boxed{(g\circ f)(x)=0} \]
and:
\[ \boxed{(f\circ g)(-3)=12,\quad (f\circ g)(5)=0,\quad (g\circ f)(-2)=0} \]
🚀 Exam Shortcut
- First convert modulus functions into piecewise form
- Notice \(g(x)\ge0\) and \(f(x)\ge0\)
- This makes second composition easier