Tag Archives: Functions
$y = f(x) = \frac{x}{{x + 5}}$ $f(5x) = g(y)$ $g(y) = ?$
$f(5x) = \frac{{5x}}{{5x + 5}} = \frac{x}{{x + 1}}$
Given, $y = \frac{x}{{x + 5}}$
$\therefore y.x + 5y = x$ Continue reading $y = f(x) = \frac{x}{{x + 5}}$ $f(5x) = g(y)$ $g(y) = ?$
f is differentiable function such that$f(x + y) – f(x) = f(y) + 2xy$$x,y \in R$$f(x) = ?$
We have, $f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) – f(x)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2xh + f(h)}}{h}$
$ = 2x + \mathop {\lim }\limits_{h \to 0} \frac{{f(h)}}{h}$ Continue reading f is differentiable function such that
$f(x + y) – f(x) = f(y) + 2xy$
$x,y \in R$
$f(x) = ?$
$f(x) + f\left( {\frac{1}{{1 – x}}} \right) = x,\forall x \in \mathbb{R}$$f(x) = ?$
$x \to 1 – \frac{1}{x}$ yields,
$f\left( {1 – \frac{1}{x}} \right) + f\left[ {\frac{1}{{1 – \left( {1 – \frac{1}{x}} \right)}}} \right] = 1 – \frac{1}{x}$
$ \Rightarrow f\left( {\frac{{x – 1}}{x}} \right) + f(x) = \frac{{x – 1}}{x}……..(A)$ Continue reading $f(x) + f\left( {\frac{1}{{1 – x}}} \right) = x,\forall x \in \mathbb{R}$
$f(x) = ?$
Find $f(x)$ such that, $f(x) + f\left( {\frac{1}{x}} \right) + 3f( – x) = x$
$x \to \frac{1}{x}$ in the original equation yields,
$f\left( {\frac{1}{x}} \right) + f(x) + 3f\left( { – \frac{1}{x}} \right) = \frac{1}{x}$ ……..(A)
Original equation – (A) yields,
$3f( – x) – 3f\left( { – \frac{1}{x}} \right) = x – \frac{1}{x}$ Continue reading Find $f(x)$ such that,
$f(x) + f\left( {\frac{1}{x}} \right) + 3f( – x) = x$
$\begin{array}{l}f\left( {x + \frac{1}{x} + 4} \right) = {x^2} + \frac{1}{{{x^2}}} + 16\\f(17) = ?\end{array}$
Using the substitution $x + \frac{1}{x} + 4 = t$, we have
$\begin{array}{l}x + \frac{1}{x} = t – 4\\ \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = {(t – 4)^2}\\ \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 16 = {(t – 4)^2} + 14\end{array}$
Now, $f(t) = {(t – 4)^2} + 14$
$\therefore f(17) = {(17 – 4)^2} + 14 = 169 + 14 = 183$
$f(x) = \frac{{{x^2} – 3x – 6}}{{{x^2} + 2x + 4}}$Domain/Range/Onto/Into
Let $f:R \to R$ be defined by $f(x) = \frac{{{x^2} – 3x – 6}}{{{x^2} + 2x + 4}}$. Then which of the following statements is(are) correct?
(A) f is onto
(B) Range of f is $\left[ { – \frac{3}{2},2} \right]$
(C) f is into
(D) Domain of f is [-4, 0] Continue reading $f(x) = \frac{{{x^2} – 3x – 6}}{{{x^2} + 2x + 4}}$
Domain/Range/Onto/Into