Find \(f \circ g\) and \(g \circ f\) for \(f(x)=\tan x\) and \(g(x)=\sqrt{1-x^2}\)
📺 Video Explanation
📝 Question
Let:
\[ f:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to\mathbb{R},\qquad f(x)=\tan x \]
and
\[ g:[-1,1]\to\mathbb{R},\qquad g(x)=\sqrt{1-x^2} \]
Find:
- \((f\circ g)(x)\)
- \((g\circ f)(x)\)
✅ Solution
🔹 Step 1: Find range of \(g(x)\)
Given:
\[ g(x)=\sqrt{1-x^2},\qquad x\in[-1,1] \]
Since:
\[ 0\le 1-x^2\le 1 \]
So:
\[ 0\le g(x)\le 1 \]
Thus:
\[ \text{Range}(g)=[0,1] \]
This lies inside the domain of \(f\), because:
\[ [0,1]\subset\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \]
So \(f\circ g\) is defined on all of \([-1,1]\). :contentReference[oaicite:1]{index=1}
🔹 Step 2: Find \((f\circ g)(x)\)
By definition:
\[ (f\circ g)(x)=f(g(x)) \]
Substitute:
\[ (f\circ g)(x)=f\left(\sqrt{1-x^2}\right) \]
Since:
\[ f(x)=\tan x \]
So:
\[ \boxed{(f\circ g)(x)=\tan\left(\sqrt{1-x^2}\right),\quad x\in[-1,1]} \]
🔹 Step 3: Find domain of \(g\circ f\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
For this to exist:
\[ f(x)=\tan x\in[-1,1] \]
That means:
\[ -1\le \tan x\le 1 \]
On:
\[ \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \]
tan x is increasing, so:
\[ -\frac{\pi}{4}\le x\le \frac{\pi}{4} \]
Hence:
\[ \text{Domain}(g\circ f)=\left[-\frac{\pi}{4},\frac{\pi}{4}\right] \]
The endpoints are valid because \(\tan(\pm \pi/4)=\pm1\), which belong to the domain of \(g\). :contentReference[oaicite:2]{index=2}
🔹 Step 4: Find \((g\circ f)(x)\)
\[ (g\circ f)(x)=g(\tan x) \]
Substitute in \(g\):
\[ (g\circ f)(x)=\sqrt{1-\tan^2 x} \]
Therefore:
\[ \boxed{(g\circ f)(x)=\sqrt{1-\tan^2 x},\quad x\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]} \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=\tan\left(\sqrt{1-x^2}\right),\quad x\in[-1,1]} \]
\[ \boxed{(g\circ f)(x)=\sqrt{1-\tan^2 x},\quad x\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]} \]
🚀 Exam Shortcut
- First check whether range of inner function fits domain of outer function
- For square root, inside must be non-negative
- For \(g\circ f\), solve \(-1\le \tan x\le 1\)