Question

find the range of f x is equal to x^8 + 2 where X is real number​

Answer 1

Answer:

[ 2 , ∞ )

Step-by-step explanation:

Given :

f ( x ) = x⁸ + 2  , x ∈ R

We have to find range of f ( x )

Let say :

y =  x⁸ + 2

= >  x⁸  = y - 2

$\displaystyle \sf \longrightarrow x = (y-2)^{1/8}$

= > y - 2 ≥ 0

= > y ≥ 2

Range of f ( x ) , y ∈ [ 2 , ∞ )

Hence we get required answer!