Question

What is the greatest value in the range of f(x) = x^2 - 3 for the domain (-3,0,1,2)

Answer 1

Answer:

f(-3)=6 is the greatest value in the range of $f(x)=x^2-3$ for the domain (-3,0,1,2)

Step-by-step explanation:

Given that the function f is defined for range by $f(x)=x^2-3$ for the domain (-3,0,1,2)

To find the greatest value in the range of $f(x)=x^2-3$ for the domain (-3,0,1,2):

$f(x)=x^2-3$ for the domain (-3,0,1,2)

That is put x=-3 in the given function $f(x)=x^2-3$ we get

$f(-3)=(-3)^2-3$

$=3^2-3$

$=9-3$

$=6$

Therefore f(-3)=6

put x=0 in the given function $f(x)=x^2-3$ we get

$f(0)=(0)^2-3$

$=0-3$

$=-3$

Therefore f(0)=-3

put x=1 in the given function $f(x)=x^2-3$ we get

$f(1)=(1)^2-3$

$=1-3$

$=-2$

Therefore f(1)=-2

put x=-3 in the given function $f(x)=x^2-3$ we get

$f(2)=(2)^2-3$

$=4-3$

$=1$

Therefore f(2)=1

Comparing the values of f(-3)=6,f(0)=-3,f(1)=-2,and f(2)=1 to find the greatest value in the range of f(x) = x^2 - 3 for the domain (-3,0,1,2) we get

Therefore f(-3)=6 is the greatest value in the range of $f(x)=x^2-3$ for the domain (-3,0,1,2)