Question

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

Answer 1

Hi

if 0 ≤ x ≤ 2

$0^2\leq x^2\leq 2^2$

Multiplied by 5..... $5×0^2 \leq 5x^2\leq 5×2^2$

add 4  .........$5×0^2+4 \leq 5x^2+4\leq 5×2^2$+4

so ........$4\leq 5x^2+4\leq 24$