April 2026

Find \(f \circ g\) and \(g \circ f\) for \(f(x)=\sqrt{1-x}\) and \(g(x)=\ln x\) 📺 Video Explanation 📝 Question Let functions \(f\) and \(g\) be defined as: \[ f(x)=\sqrt{1-x},\qquad g(x)=\ln x \] Find: \((f\circ g)(x)\) \((g\circ f)(x)\) ✅ Solution 🔹 Find \((f\circ g)(x)\) By definition: \[ (f\circ g)(x)=f(g(x)) \] Substitute \(g(x)=\ln x\): \[ (f\circ g)(x)=f(\ln x)=\sqrt{1-\ln […]

Prove \(g \circ f=f+f\) for Any Real Function \(f\) and \(g(x)=2x\) 📺 Video Explanation 📝 Question Let \(f\) be any real-valued function and let: \[ g(x)=2x \] Prove that: \[ g\circ f=f+f \] ✅ Solution 🔹 Step 1: Write the composite function By definition of composition: \[ (g\circ f)(x)=g(f(x)) \] 🔹 Step 2: Substitute \(g(x)=2x\)

Prove \(f \circ g=g\circ(fh)\) for \(f(x)=\sin x,\ g(x)=2x,\ h(x)=\cos x\) 📺 Video Explanation 📝 Question Let real functions be defined as: \[ f(x)=\sin x,\qquad g(x)=2x,\qquad h(x)=\cos x \] Prove that: \[ f\circ g=g\circ(fh) \] where: \[ (fh)(x)=f(x)\cdot h(x) \] ✅ Solution 🔹 Step 1: Find \((f\circ g)(x)\) By definition: \[ (f\circ g)(x)=f(g(x)) \] Substitute \(g(x)=2x\):

Find \(g \circ f\) and \(f \circ g\) for \(f(x)=\sin x\) and \(g(x)=2x\) 📺 Video Explanation 📝 Question Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as: \[ f(x)=\sin x,\qquad g(x)=2x \] Find: \((g\circ f)(x)\) \((f\circ g)(x)\) Also, check whether these are equal functions. ✅ Solution 🔹 Find \((g\circ f)(x)\) By definition: \[ (g\circ f)(x)=g(f(x)) \]

Find \((f \circ g)\), \((g \circ f)\), \((f \circ f)\), and \(f^2\) for \(f(x)=2x+5\) and \(g(x)=x^2+1\) 📺 Video Explanation 📝 Question Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as: \[ f(x)=2x+5,\qquad g(x)=x^2+1 \] Find: \((f\circ g)(x)\) \((g\circ f)(x)\) \((f\circ f)(x)\) \(f^2(x)\) ✅ Solution 🔹 (i) Find \((f\circ g)(x)\) By definition: \[ (f\circ g)(x)=f(g(x)) \] Substitute

Prove \(f \circ f=f\) for \(f(x)=|x|\) 📺 Video Explanation 📝 Question If: \[ f(x)=|x| \] prove that: \[ f\circ f=f \] ✅ Solution 🔹 Step 1: Find \((f\circ f)(x)\) By definition: \[ (f\circ f)(x)=f(f(x)) \] Since: \[ f(x)=|x| \] Substitute: \[ (f\circ f)(x)=f(|x|) \] Again using \(f(t)=|t|\): \[ (f\circ f)(x)=||x|| \] 🔹 Step 2: Use

Show \(f \circ g \ne g \circ f\) for \(f(x)=x^2+x+1\) and \(g(x)=\sin x\) 📺 Video Explanation 📝 Question Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as: \[ f(x)=x^2+x+1,\qquad g(x)=\sin x \] Show that: \[ f\circ g \ne g\circ f \] ✅ Solution 🔹 Find \((f\circ g)(x)\) By definition: \[ (f\circ g)(x)=f(g(x)) \] Substitute \(g(x)=\sin x\):

Find \(f \circ g\) and \(g \circ f\) for \(f(x)=x^2+2\) and \(g(x)=1-\frac{1}{1-x}\) 📺 Video Explanation 📝 Question Let: \[ f(x)=x^2+2,\qquad g(x)=1-\frac{1}{1-x} \] Find: \((f\circ g)(x)\) \((g\circ f)(x)\) ✅ Solution 🔹 Find \((f\circ g)(x)\) By definition: \[ (f\circ g)(x)=f(g(x)) \] Substitute \(g(x)\): \[ (f\circ g)(x)=\left(1-\frac{1}{1-x}\right)^2+2 \] Simplify inside: \[ 1-\frac{1}{1-x}=\frac{(1-x)-1}{1-x}=\frac{-x}{1-x}=\frac{x}{x-1} \] So: \[ (f\circ g)(x)=\left(\frac{x}{x-1}\right)^2+2 \]

Find \(f \circ g\) and \(g \circ f\) for Constant Function \(f(x)=c\) and \(g(x)=\sin x^2\) 📺 Video Explanation 📝 Question Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as: \[ f(x)=c,\quad c\in\mathbb{R} \] and \[ g(x)=\sin x^2 \] Find: \((f\circ g)(x)\) \((g\circ f)(x)\) ✅ Solution 🔹 Find \((f\circ g)(x)\) By definition: \[ (f\circ g)(x)=f(g(x)) \] Substitute:

Find \(f \circ g\) and \(g \circ f\) for \(f(x)=x+1\) and \(g(x)=2x+3\) 📺 Video Explanation 📝 Question Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as: \[ f(x)=x+1,\qquad g(x)=2x+3 \] Find: \((f\circ g)(x)\) \((g\circ f)(x)\) ✅ Solution 🔹 Find \((f\circ g)(x)\) By definition: \[ (f\circ g)(x)=f(g(x)) \] Substitute \(g(x)=2x+3\): \[ (f\circ g)(x)=f(2x+3) \] Since: \[ f(x)=x+1