Respuesta :
Range of [tex]f(x)=(x + 3)^2 + 4[/tex] is [4,∞).
Step-by-step explanation:
Here we have , function [tex]f(x)=(x + 3)^2 + 4[/tex] , We need to find range of this function . Let's find out:
In the function [tex]f(x)=(x + 3)^2 + 4[/tex] , Let's focus on [tex](x+3)^2[/tex] : It's a perfect square whose value will be always positive and greater then equal to zero i.e.
⇒ [tex](x+3)^2\geq 0[/tex] , So minimum value of [tex](x+3)^2[/tex] is 0 at x=-3 . ∴
⇒ [tex]f(x)=(x + 3)^2 + 4[/tex]
⇒ [tex]f(-3)=(-3+ 3)^2 + 4[/tex]
⇒ [tex]f(-3)= 4[/tex]
Minimum value of [tex]f(x)=(x + 3)^2 + 4[/tex] is 4 , And maximum value of [tex]f(x)=(x + 3)^2 + 4[/tex] can be ∞ as value of x can be increased . Therefore , Range of [tex]f(x)=(x + 3)^2 + 4[/tex] is [4,∞).
