If f: R → R be a mapping defined by f(x) = x3 + 5, then f-1(x) is equal to:

Question

If f: R → R be a mapping defined by f(x) = x3 + 5, then f-1(x) is equal to: (A) (x + 3)1/3 (B) (x - 5)1/3 (C) (5 - x)1/3 (D) (5-x)

Tardigrade Admin · Accepted Answer

Let f: R → R defined as f (x) = x3 + 5 Let f (x) = y ⇒ x = f-1(y) .....(i) Given: f(x) = x3 + 5 ⇒ y = x3 + 5 ⇒ y - 5 = x3 ⇒ (y - 5)(1/3) = x = f-1 (y) (from (i)) ∴ f-1 (x) = (x - 5)(1/3)

Q. If $f : R \to R$ be a mapping defined by $f (x) = x^{3} + 5$ , then $f^{- 1} (x)$ is equal to:

$(x + 3)^{1/3}$

$(x - 5)^{1/3}$

$(5 - x)^{1/3}$

(5-x)

Let f : R $\to$ R defined as $f (x) = x^{3} + 5$
Let f (x) = y $\Rightarrow$ x = $f^{- 1}$ (y) .....(i)
Given : $f (x) = x^{3} + 5$
$\Rightarrow y = x^{3} + 5$
$\Rightarrow y - 5 = x^{3}$
$\Rightarrow (y - 5)^{\frac{1}{3}} = x = f^{- 1} (y)$ (from (i))
$∴ f^{- 1} (x) = (x - 5)^{\frac{1}{3}}$

Q. If f:R→R be a mapping defined by f(x)=x3+5, then f−1(x) is equal to:

(x+3)1/3

(x−5)1/3

(5−x)1/3