April 2026

Verify \(h\circ(g\circ f)=(h\circ g)\circ f\) for Given Functions 📺 Video Explanation 📝 Question Let: \[ f:\mathbb{N}\to\mathbb{N},\qquad f(x)=2x \] \[ g:\mathbb{N}\to\mathbb{N},\qquad g(y)=3y+4 \] \[ h:\mathbb{N}\to\mathbb{R},\qquad h(z)=\sin z \] Show that: \[ h\circ(g\circ f)=(h\circ g)\circ f \] ✅ Solution 🔹 Step 1: Find \(g\circ f\) By definition: \[ (g\circ f)(x)=g(f(x)) \] Substitute \(f(x)=2x\): \[ (g\circ f)(x)=g(2x)=3(2x)+4 \] […]

Verify Associativity of Composite Functions for Given Mappings 📺 Video Explanation 📝 Question Let: \[ f:\mathbb{N}\to \mathbb{Z}_0,\qquad f(x)=2x \] \[ g:\mathbb{Z}_0\to \mathbb{Q},\qquad g(x)=\frac{1}{x} \] \[ h:\mathbb{Q}\to \mathbb{R},\qquad h(x)=e^x \] Verify associativity: \[ h\circ(g\circ f)=(h\circ g)\circ f \] ✅ Solution 🔹 Step 1: Check that compositions are defined Since: \(f:\mathbb{N}\to \mathbb{Z}_0\) \(g:\mathbb{Z}_0\to \mathbb{Q}\) \(h:\mathbb{Q}\to \mathbb{R}\) Both:

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

Show \(f \circ g \ne g \circ f\) for \(f(x)=x^2\) and \(g(x)=x+1\) 📺 Video Explanation 📝 Question Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined by: \[ f(x)=x^2,\qquad g(x)=x+1 \] 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 \(g(x)=x+1\): \[ (f\circ g)(x)=f(x+1)

Find \(f \circ g\) and \(g \circ f\) for \(f(x)=x^2\) and \(g(x)=\sqrt{x}\) on \(\mathbb{R}_+\) 📺 Video Explanation 📝 Question Let \(\mathbb{R}_+\) be the set of all non-negative real numbers. If functions \(f:\mathbb{R}_+\to\mathbb{R}_+\) and \(g:\mathbb{R}_+\to\mathbb{R}_+\) are defined by: \[ f(x)=x^2,\qquad g(x)=\sqrt{x} \] Find: \((f\circ g)(x)\) \((g\circ f)(x)\) Also, check whether they are equal functions. ✅ Solution

Find \((f \circ g)(2)\) and \((g \circ f)(1)\) for \(f(x)=x^2+8\) and \(g(x)=3x^3+1\) 📺 Video Explanation 📝 Question Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as: \[ f(x)=x^2+8,\qquad g(x)=3x^3+1 \] Find: \((f\circ g)(2)\) \((g\circ f)(1)\) ✅ Solution 🔹 Find \((f\circ g)(2)\) By definition: \[ (f\circ g)(2)=f(g(2)) \] First find: \[ g(2)=3(2)^3+1 \] \[ =3(8)+1=25 \] Now:

Verify \(f\) and \(g\) are Inverse Functions Using \(f \circ g\) and \(g \circ f\) 📺 Video Explanation 📝 Question Let \[ A=\{a,b,c\},\qquad B=\{u,v,w\} \] Let: \[ f=\{(a,v),(b,u),(c,w)\} \] and \[ g=\{(u,b),(v,a),(w,c)\} \] where \(f:A\to B\) and \(g:B\to A\). Verify whether \(f\) and \(g\) are inverse functions. ✅ Solution 🔹 Step 1: Find \(g\circ f\)

Show \(g \circ f\) is Defined but \(f \circ g\) is Not Defined 📺 Video Explanation 📝 Question Let \[ f=\{(1,-1),(4,-2),(9,-3),(16,4)\} \] and \[ g=\{(-1,-2),(-2,-4),(-3,-6),(4,8)\} \] Show that \(g\circ f\) is defined while \(f\circ g\) is not defined. Also, find \(g\circ f\). ✅ Solution 🔹 Step 1: Check whether \(g\circ f\) is defined For \(g\circ

Find \(g \circ f\) and \(f \circ g\) for Given Relations 📺 Video Explanation 📝 Question Let \[ f=\{(3,1),(9,3),(12,4)\} \] and \[ g=\{(1,3),(3,3),(4,9),(5,9)\} \] Show that \(g\circ f\) and \(f\circ g\) are both defined. Also, find \(f\circ g\) and \(g\circ f\). ✅ Solution 🔹 Step 1: Check whether \(g\circ f\) is defined For \(g\circ f\)

Find \(g \circ f\) and \(f \circ g\) for \(f(x)=8x^3\) and \(g(x)=\sqrt[3]{x}\) 📺 Video Explanation 📝 Question Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as: \[ f(x)=8x^3,\qquad g(x)=\sqrt[3]{x} \] Find: \((g\circ f)(x)\) \((f\circ g)(x)\) ✅ Solution 🔹 Find \((g\circ f)(x)\) By definition: \[ (g\circ f)(x)=g(f(x)) \] Substitute \(f(x)=8x^3\): \[ g(8x^3)=\sqrt[3]{8x^3} \] Now: \[ \sqrt[3]{8}=2,\qquad \sqrt[3]{x^3}=x