If the function f:[1, ∞) arrow[1, ∞) is defined by f(x)=2x(x-1) then f-1(x) is

Question

If the function f:[1, ∞) arrow[1, ∞) is defined by f(x)=2x(x-1) then f-1(x) is (A) (1 / 2)x(x-1) (B) (1/2)(1+√1+4 log 2 x) (C) (1/2)(1-√1+4 log 2 x) (D) not defined

Tardigrade Admin · Accepted Answer

Let f(x)=y=2x(x-1) ⇒ log y=x(x-1) log 2 ⇒ x2-x- log 2 y=0 ⇒ x=(1 ± √1+4 log 2 y/2) ⇒ x=f-1(y)=(1 ± √1+4 log 2 y/2) ⇒ f-1(x)=(1+√1+4 log 2 y/2) text as f-1(x) ≥ 1 .

Q. If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ then $f^{- 1} (x)$ is

$(1/2)^{x (x - 1)}$

$\frac{1}{2} (1 + 1 + 4 lo g_{2} x)$

$\frac{1}{2} (1 - 1 + 4 lo g_{2} x)$

not defined

Q. If the function f:[1,∞)→[1,∞) is defined by f(x)=2x(x−1) then f−1(x) is

(1/2)x(x−1)

21​(1+1+4log2​x​)

21​(1−1+4log2​x​)

not defined

Let f(x)=y=2x(x−1) ⇒logy=x(x−1)log2 ⇒x2−x−log2​y=0 ⇒x=21±1+4log2​y​​ ⇒x=f−1(y)=21±1+4log2​y​​ ⇒f−1(x)=21+1+4log2​y​​ as f−1(x)≥1.

Q. If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ then $f^{- 1} (x)$ is

$(1/2)^{x (x - 1)}$

$\frac{1}{2} (1 + 1 + 4 lo g_{2} x)$

$\frac{1}{2} (1 - 1 + 4 lo g_{2} x)$