Consider f- R- -4- - given by fx - x2 - 4- Show that f is invertible with the inverse f-1 of f given by f-1y - -y - 4- where R- is the set of all non-negative real numbers-

f: R _ 1 #160;→ ( 4 , ∞ #160; ) f( x ) = x ^ 2 +4 f( x _ 1 ) =f( x _ 2 ) #160; x _ 1 ^ 2 +4= x _ 2 ^ 2 +4 x _ 1 = x _ 2 #16 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Properties of Inverse Function

Consider f:...

Question

Consider $f : R_{+} \to [4, \infty)$ given by $f (x) = x^{2} + 4$ . Show that $f$ is invertible with the inverse $f^{- 1}$ of $f$ given by $f^{- 1} (y) = \sqrt{y - 4}$ , where $R_{+}$ is the set of all non-negative real numbers.

Open in App

Solution

$f : R_{1} \to (4, \infty)$
$f (x) = x^{2} + 4$
$f (x_{1}) = f (x_{2})$
${x_{1}}^{2} + 4 = {x_{2}}^{2} + 4$
$x_{1} = x_{2}$ $f (x)$ is on-one.
$f (x) = x^{2} + 4$
$y = x^{2} + 4$
$x^{2} = y - 4$
$x = \sqrt{y - 4}$
$f (x) = x^{2} + 4$
$f (x) = {(\sqrt{y - 4})}^{2} + 4$
$f (x) = y - 4 + 4$
$f (x) = y$
$f (x)$ is onto function.
So, $f (x)$ is bijective, Hence it is invertible
$f (x) = x^{2} + 4$
$y = x^{2} + 4$
$y = x^{2} - 4$
$y = \sqrt{x^{2} - 4}$
$f^{- 1} (x) = \sqrt{x - 4}$

Suggest Corrections

Similar questions

Q. Consider

f : R + \to [4, \infty]

given by

f (x) = x^{2} + 4

. Show
that

f

is invertible with the inverse

f^{- 1}

f

given by

f^{- 1} (y) = \sqrt{y - 4}

where

R_{+}

is the set of all non-negative real numbers.

Q. Consider f : R + → [4, ∞ ) given by f ( x ) = x 2 + 4. Show that f is invertible with the inverse f −1 of given f by , where R + is the set of all non-negative real numbers.

Q. Let

R_{+}

be the set of all non-negative real numbers. Show that the function

f : R_{+} \to [4, \infty)

given by

f (x) = x^{2} + 4

is invertible and write the inverse of

f

Consider $f : R_{+} \to [4, \infty)$ given by $f (x) = x^{2} + 4$ . Show that f is ivnertible with the inverse $f^{- 1}$ given by $f^{- 1} (y) = \sqrt{y - 4}$ . where $R_{+}$ is the set of all non-negative real numbers.

Q. Consider f : R → R₊ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with inverse f⁻¹ of f given by f⁻¹

(x) = \sqrt{x - 4}

, where R⁺ is the set of all non-negative real numbers.

Join BYJU'S Learning Program

Consider f:R+→[4,∞) given by f(x)=x2+4. Show that f is invertible with the inverse f−1 of f given by f−1(y)=√y−4, where R+ is the set of all non-negative real numbers.

f:R1→(4,∞)f(x)=x2+4f(x1)=f(x2) x12+4=x22+4x1=x2 f(x) is on-one.f(x)=x2+4y=x2+4x2=y−4x=√y−4 f(x)=x2+4f(x)=(√y−4)2+4f(x)=y−4+4f(x)=yf(x) is onto function.So,f(x) is bijective, Hence it is invertiblef(x)=x2+4y=x2+4y=x2−4y=√x2−4f−1(x)=√x−4

Consider $f : R_{+} \to [4, \infty)$ given by $f (x) = x^{2} + 4$ . Show that $f$ is invertible with the inverse $f^{- 1}$ of $f$ given by $f^{- 1} (y) = \sqrt{y - 4}$ , where $R_{+}$ is the set of all non-negative real numbers.