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What is the domain of the function: $f(x) =sqrt(x^2(x-3)(x-4))$ ?

2 Answers

Oct 2, 2015

$D_(f(x)) = (-oo,3]uu[4,+oo)$

Explanation:

Given
$color(white)("XXX")f(x)=sqrt(x^2(x-3)(x-4))$

To find the domain we need to determine which values of $x$ are not valid.

Since the $sqrt("negative value")$ is undefined (for Real numbers)

$x^2(x-3)(x-4) >= 0$

$x^2 >= 0$ for all $x in RR$
$(x-3) > 0$ for all $x>3, in RR$
$(x-4) > 0$ for all $x>4, in RR$

The only combination for which
$color(white)("XXX")x^2(x-3)(x-4) < 0$
is when $(x-3) > 0$ and $(x-4) < 0$

That is the only non-valid values for (Real) $x$ occur when
$color(white)("XXX")x > 3$ and $x < 4$

Oct 2, 2015

$(-oo, 3] uu [4, oo)$

Explanation:

The domain is where the radicand (the expression under the square root sign) is non-negative.

We know that $x^2 >= 0$ for all $x in RR$ .

So in order that $x^2(x-3)(x-4) >= 0$ , we must either have $x^2 = 0$ or $(x-3)(x-4) >= 0$ .

When $x<=3$ , both $(x-3) <= 0$ and $(x-4)<=0$ , so $(x-3)(x-4) >= 0$

When $3 < x < 4$ , $(x-3) > 0$ and $(x-4) < 0$ , so $(x-3)(x-4) < 0$ .

When $x >= 4$ , both $(x-3)>=0$ and $(x-4)>=0$ , so $(x-3)(x-4) >= 0$ .

So $x^2(x-3)(x-4)>=0$ when $x in (-oo,3] uu [4, oo)$

Note that this domain already includes the point $x = 0$ , so the $x^2 = 0$ condition gives us no extra points for the domain.

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