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, $$

Show that f(x)+f(1/x)=0 Show that \(f(x)+f\left(\frac1x\right)=0\) Question: If $$ f(x)=x^3-\frac1{x^3} $$ show that $$ f(x)+f\left(\frac1x\right)=0 $$ Solution Given: $$ f(x)=x^3-\frac1{x^3} $$ Replace \(x\) by \(\frac1x\): $$ f\left(\frac1x\right) = \left(\frac1x\right)^3-\frac1{\left(\frac1x\right)^3} $$ $$ = \frac1{x^3}-x^3 $$ Now, $$ f(x)+f\left(\frac1x\right) = \left(x^3-\frac1{x^3}\right) + \left(\frac1{x^3}-x^3\right) $$ $$ =0 $$ Hence, $$ \boxed{f(x)+f\left(\frac1x\right)=0} $$ “` Next Question / Full Exercise

Evaluate Piecewise Function Values Evaluate Piecewise Function Values Question: If $$ f(x)= \begin{cases} x^2, & x

Show that f(f(x)) = x Show that \(f(f(x))=x\) Question: If $$ f(x)=\frac{x+1}{x-1} $$ show that $$ f(f(x))=x $$ Solution Given: $$ f(x)=\frac{x+1}{x-1} $$ Now, $$ f(f(x)) = \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} $$ Simplify the numerator: $$ \frac{x+1}{x-1}+1 = \frac{x+1+x-1}{x-1} = \frac{2x}{x-1} $$ Simplify the denominator: $$ \frac{x+1}{x-1}-1 = \frac{x+1-x+1}{x-1} = \frac{2}{x-1} $$ Therefore, $$ f(f(x)) = \frac{\frac{2x}{x-1}}{\frac{2}{x-1}} $$

Show that f(f(f(x))) = x Show that \(f(f(f(x)))=x\) Question: If $$ f(x)=\frac{1}{1-x} $$ show that $$ f(f(f(x)))=x $$ Solution Given: $$ f(x)=\frac{1}{1-x} $$ First find \(f(f(x))\): $$ f(f(x)) = \frac{1}{1-\frac{1}{1-x}} $$ $$ = \frac{1}{\frac{(1-x)-1}{1-x}} $$ $$ = \frac{1-x}{-x} $$ $$ = \frac{x-1}{x} $$ Now find \(f(f(f(x)))\): $$ f\left(\frac{x-1}{x}\right) = \frac{1}{1-\frac{x-1}{x}} $$ $$ = \frac{1}{\frac{x-(x-1)}{x}} $$

Show that x = f(y) Show that \(x=f(y)\) Question: If $$ y=f(x)=\frac{ax-b}{bx-a} $$ show that $$ x=f(y) $$ Solution Given: $$ y=\frac{ax-b}{bx-a} $$ Now, $$ f(y)=\frac{ay-b}{by-a} $$ Substitute the value of \(y\): $$ f(y)= \frac{ a\left(\frac{ax-b}{bx-a}\right)-b }{ b\left(\frac{ax-b}{bx-a}\right)-a } $$ Taking LCM in numerator and denominator: $$ = \frac{ \frac{a(ax-b)-b(bx-a)}{bx-a} }{ \frac{b(ax-b)-a(bx-a)}{bx-a} } $$ $$

Find f(a+b) Find \(f(a+b)\) Question: If $$ f(x)=(x-a)^2(x-b)^2 $$ find $$ f(a+b) $$ Solution Given: $$ f(x)=(x-a)^2(x-b)^2 $$ Put \(x=a+b\): $$ f(a+b)=((a+b)-a)^2((a+b)-b)^2 $$ $$ =(b)^2(a)^2 $$ $$ =a^2b^2 $$ Hence, $$ \boxed{f(a+b)=a^2b^2} $$ “` Next Question / Full Exercise

Solve f(x)=f(2x+1) Solve \(f(x)=f(2x+1)\) Question: If $$ f(x)=x^2-3x+4 $$ then find the value of \(x\) satisfying $$ f(x)=f(2x+1) $$ Solution Given: $$ f(x)=x^2-3x+4 $$ First find \(f(2x+1)\): $$ f(2x+1)=(2x+1)^2-3(2x+1)+4 $$ $$ =4x^2+4x+1-6x-3+4 $$ $$ =4x^2-2x+2 $$ Now, $$ f(x)=f(2x+1) $$ $$ x^2-3x+4=4x^2-2x+2 $$ $$ 3x^2+x-2=0 $$ $$ 3x^2+3x-2x-2=0 $$ $$ 3x(x+1)-2(x+1)=0 $$ $$ (x+1)(3x-2)=0 $$

Express Function as Set of Ordered Pairs Express Function as Set of Ordered Pairs Question: Express the function $$ f:X\to\mathbb{R} $$ given by $$ f(x)=x^3+1 $$ as a set of ordered pairs, where $$ X=\{-1,0,3,9,7\} $$ Solution Given: $$ f(x)=x^3+1 $$ Find the value of \(f(x)\) for each element of \(X\). \(x\) \(f(x)=x^3+1\) \(-1\) \((-1)^3+1=0\)

Find Difference Quotient for f(x)=x² Find Difference Quotient for \(f(x)=x^2\) Question: If $$ f(x)=x^2 $$ find $$ \frac{f(1.1)-f(1)}{(1.1)-1} $$ Solution Given: $$ f(x)=x^2 $$ $$ f(1.1)=(1.1)^2=1.21 $$ $$ f(1)=1^2=1 $$ Therefore, $$ \frac{f(1.1)-f(1)}{(1.1)-1} = \frac{1.21-1}{0.1} $$ $$ = \frac{0.21}{0.1} =2.1 $$ Hence, $$ \boxed{2.1} $$ “` Next Question / Full Exercise