How do I find the domain of f(x)=5x^2+4. When 0<=x<=2
#54

Hi
if 0 ≤ x ≤ 2
[tex]0^2\leq x^2\leq 2^2[/tex]
Multiplied by 5..... [tex]5×0^2 \leq 5x^2\leq 5×2^2[/tex]
add 4 .........[tex]5×0^2+4 \leq 5x^2+4\leq 5×2^2[/tex]+4
so ........[tex]4\leq 5x^2+4\leq 24[/tex]