Question

find the domain of $f(x) = \sqrt{2x^2 - 50}$. \
ewline options: \
ewline $\circ$ $x \geq 5$ \
ewline $\circ$ $x \geq 5$ or $x \leq -5$ \
ewline $\circ$ all real numbers \
ewline $\circ$ $x \geq 25$ \
ewline the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer.

Explanation:

Step1: Set the radicand non - negative

For the function \(f(x)=\sqrt{2x^{2}-50}\), the radicand \(2x^{2}-50\) must be greater than or equal to 0. So we set up the inequality \(2x^{2}-50\geq0\).

Step2: Simplify the inequality

First, divide both sides of the inequality \(2x^{2}-50\geq0\) by 2. We get \(x^{2}-25\geq0\). Then, factor the left - hand side using the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), where \(a = x\) and \(b = 5\). So \(x^{2}-25=(x + 5)(x - 5)\geq0\).
To solve the inequality \((x + 5)(x - 5)\geq0\), we consider the critical points \(x=-5\) and \(x = 5\). These points divide the number line into three intervals: \((-\infty,-5)\), \((-5,5)\), and \((5,\infty)\).

• For the interval \((-\infty,-5)\), let's take a test point \(x=-6\). Then \((-6 + 5)(-6 - 5)=(-1)\times(-11)=11\geq0\), so the inequality holds in this interval.
• For the interval \((-5,5)\), let's take a test point \(x = 0\). Then \((0 + 5)(0 - 5)=(5)\times(-5)=-25<0\), so the inequality does not hold in this interval.
• For the interval \((5,\infty)\), let's take a test point \(x = 6\). Then \((6 + 5)(6 - 5)=(11)\times(1)=11\geq0\), so the inequality holds in this interval.

Also, when \(x=-5\) or \(x = 5\), \((x + 5)(x - 5)=0\), which satisfies the inequality \((x + 5)(x - 5)\geq0\).

So the solution to the inequality \(x^{2}-25\geq0\) is \(x\geq5\) or \(x\leq - 5\).

Answer:

\(x\geq5\) or \(x\leq - 5\) (the option with the text "x ≥ 5 or x ≤ -5")