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Polynomial Functions

If f:R - 0 ...

Question

If $f : R - {0} \to R$ is defined by $f (x) = x^{3} - \frac{1}{x^{3}}$ then show that $f (x) + f (\frac{1}{x}) = 0$ .

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Solution

$f (x) = x^{3} - \frac{1}{x^{3}}$
$f (\frac{1}{x}) = \frac{1}{x^{3}} - x^{3}$
$f (x) + f (\frac{1}{x}) = (x^{3} - \frac{1}{x^{3}}) + (\frac{1}{x^{3}} - x^{3})$
$= 0$

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Similar questions

If $f (x) = x^{3} - \frac{1}{x^{3}}$ , show that $f (x) + f (\frac{1}{x} = 0)$ .

f (x) = x^{3} - \frac{1}{x^{3}}

f (x) + f (\frac{1}{x}) = 0

f : R - {1} \to R

f (x) = \frac{x + 1}{x - 1}

f (x) + f (\frac{1}{x}) = 0

(x \neq 0)

Q. Classify the following functions as injection, surjection or bijection :
(i) f : N → N given by f(x) = x²
(ii) f : Z → Z given by f(x) = x²
(iii) f : N → N given by f(x) = x³
(iv) f : Z → Z given by f(x) = x³
(v) f : R → R, defined by f(x) = |x|
(vi) f : Z → Z, defined by f(x) = x² + x
(vii) f : Z → Z, defined by f(x) = x − 5
(viii) f : R → R, defined by f(x) = sinx
(ix) f : R → R, defined by f(x) = x³ + 1
(x) f : R → R, defined by f(x) = x³ − x
(xi) f : R → R, defined by f(x) = sin²x + cos²x
(xii) f : Q − {3} → Q, defined by

f (x) = \frac{2 x + 3}{x - 3}

(xiii) f : Q → Q, defined by f(x) = x³ + 1
(xiv) f : R → R, defined by f(x) = 5x³ + 4
(xv) f : R → R, defined by f(x) = 3 − 4x
(xvi) f : R → R, defined by f(x) = 1 + x²
(xvii) f : R → R, defined by f(x) =

\frac{x}{x^{2} + 1}

[NCERT EXEMPLAR]

f : R \to R

f (x) = x^{3}

f^{- 1} (8) =

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