Find the Domain of f(x)=√(x-2) Find the Domain of \(f(x)=\sqrt{x-2}\) Question: Find the domain of the following real valued function of real variable: $$ f(x)=\sqrt{x-2} $$ Solution Given: $$ f(x)=\sqrt{x-2} $$ For a square root function, the expression inside the root must be non-negative. Therefore, $$ x-2\ge0 $$ $$ x\ge2 $$ Hence, the domain is: […]

Find the Domain of f(x)=(x²+2x+1)/(x²-8x+12) Find the Domain of \(f(x)=\frac{x^2+2x+1}{x^2-8x+12}\) Question: Find the domain of the following real valued function of real variable: $$ f(x)=\frac{x^2+2x+1}{x^2-8x+12} $$ Solution Given: $$ f(x)=\frac{x^2+2x+1}{x^2-8x+12} $$ The denominator cannot be zero. $$ x^2-8x+12\ne0 $$ Factorizing: $$ (x-6)(x-2)\ne0 $$ Therefore, $$ x\ne6 \quad \text{and} \quad x\ne2 $$ Hence, all real numbers

Find the Domain of f(x)=(2x+1)/(x²-9) Find the Domain of \(f(x)=\frac{2x+1}{x^2-9}\) Question: Find the domain of the following real valued function of real variable: $$ f(x)=\frac{2x+1}{x^2-9} $$ Solution Given: $$ f(x)=\frac{2x+1}{x^2-9} $$ The denominator cannot be zero. $$ x^2-9\ne0 $$ $$ (x-3)(x+3)\ne0 $$ Therefore, $$ x\ne3 \quad \text{and} \quad x\ne-3 $$ Hence, all real numbers except

Find the Domain of f(x)=(3x-2)/(x+1) Find the Domain of \(f(x)=\frac{3x-2}{x+1}\) Question: Find the domain of the following real valued function of real variable: $$ f(x)=\frac{3x-2}{x+1} $$ Solution Given: $$ f(x)=\frac{3x-2}{x+1} $$ The denominator cannot be zero. Therefore, $$ x+1\ne0 $$ $$ x\ne-1 $$ Hence, all real numbers except \(-1\) are allowed. Therefore, the domain is:

Find the Domain of f(x)=1/(x-7) Find the Domain of \(f(x)=\frac1{x-7}\) Question: Find the domain of the following real valued function of real variable: $$ f(x)=\frac1{x-7} $$ Solution Given: $$ f(x)=\frac1{x-7} $$ The denominator cannot be zero. Therefore, $$ x-7\ne0 $$ $$ x\ne7 $$ Hence, all real numbers except \(7\) are allowed. Therefore, the domain is:

Find the Domain of f(x)=1/x Find the Domain of \(f(x)=\frac1x\) Question: Find the domain of the following real valued function of real variable: $$ f(x)=\frac1x $$ Solution Given: $$ f(x)=\frac1x $$ The denominator of a fraction cannot be zero. Therefore, $$ x\ne0 $$ Hence, all real numbers except \(0\) are allowed. Therefore, the domain is:

Find f(x) from Functional Equation Find \(f(x)\) from Functional Equation Question: If for non-zero \(x\), $$ af(x)+bf\left(\frac1x\right)=\frac1x-5, $$ where $$ a\ne b, $$ then find \(f(x)\). Solution Given: $$ af(x)+bf\left(\frac1x\right)=\frac1x-5 \quad \cdots (1) $$ Replace \(x\) by \(\frac1x\): $$ af\left(\frac1x\right)+bf(x)=x-5 \quad \cdots (2) $$ Multiply equation (1) by \(a\): $$ a^2f(x)+abf\left(\frac1x\right)=\frac{a}{x}-5a \quad \cdots (3) $$

Prove Identities for f(x) Prove Identities for \(f(x)\) Question: If $$ f(x)=\frac{x-1}{x+1}, $$ then show that: (i) $$ f\left(\frac1x\right)=-f(x) $$ (ii) $$ f\left(-\frac1x\right)=-\frac1{f(x)} $$ Solution (i) Show that \(f\left(\frac1x\right)=-f(x)\) Given: $$ f(x)=\frac{x-1}{x+1} $$ Replace \(x\) by \(\frac1x\): $$ f\left(\frac1x\right) = \frac{\frac1x-1}{\frac1x+1} $$ $$ = \frac{\frac{1-x}{x}}{\frac{1+x}{x}} $$ $$ = \frac{1-x}{1+x} $$ $$ = -\frac{x-1}{x+1} $$ $$

Prove that f(f(x)) = x Prove that \(f(f(x))=x\) Question: If $$ f(x)=(a-x^n)^{1/n}, $$ where $$ a>0 \text{ and } n\in\mathbb N, $$ then prove that $$ f(f(x))=x $$ for all \(x\). Solution Given: $$ f(x)=(a-x^n)^{1/n} $$ Now, $$ f(f(x)) = \left[a-\left((a-x^n)^{1/n}\right)^n\right]^{1/n} $$ Since $$ \left((a-x^n)^{1/n}\right)^n=a-x^n $$ Therefore, $$ f(f(x)) = \left[a-(a-x^n)\right]^{1/n} $$ $$ = (x^n)^{1/n}

Show that f(tanθ)=sin2θ Show that \(f(\tan\theta)=\sin2\theta\) Question: If $$ f(x)=\frac{2x}{1+x^2} $$ show that $$ f(\tan\theta)=\sin2\theta $$ Solution Given: $$ f(x)=\frac{2x}{1+x^2} $$ Put \(x=\tan\theta\): $$ f(\tan\theta) = \frac{2\tan\theta}{1+\tan^2\theta} $$ Using $$ 1+\tan^2\theta=\sec^2\theta $$ $$ f(\tan\theta) = \frac{2\tan\theta}{\sec^2\theta} $$ $$ = 2\tan\theta\cos^2\theta $$ $$ = 2\left(\frac{\sin\theta}{\cos\theta}\right)\cos^2\theta $$ $$ = 2\sin\theta\cos\theta $$ Using $$ \sin2\theta=2\sin\theta\cos\theta $$ Therefore, $$