Question

Domain for g(x) = √4x – x^2​

Answer 1

Answer:

$\large\boxed{0\leq x\leq4\to x\in[0,\ 4]}$

Step-by-step explanation:

We know: √x exist if x ≥ 0.

We have $g(x)=\sqrt{4x-x^2}$.

The domain:

$4x-x^2\geq0\\\\x(4-x)\geq0$

Find the zeros of the equation

$x(4-x)=0\iff x=0\ or\ 4-x=0\\\\x=0\ or\ x=4$

$ax^2+bx+c=-x^2+4x\to a=-1$

the parabola open down.

Look at the picture.

$x\geq0\ \wedge\ x\leq4\to0\leq x\leq4\to x\in[0,\ 4]$

Answer 2

Answer:

The domain is 0 ≤ x ≤ 4,

or in interval notation it is [0, 4].

Step-by-step explanation:

g(x) = √(4x – x^2)

4x - x^2 cannot have a negative value because of the square root sign.

4x - x^2 = 0

x(4 - x) = 0

x = 0 , 4.

The highest value is 4 and the lowest is 0 . Values in between like 1 are in the domain ( for example  √(4(1) - 1) = √3).

x has to be  between 0 and 4 inclusive.