which statements are true for the functions $g(x)=x^{2}$ and $h(x)=-x^{2}$? check all that apply.\nfor any…

Answer:

• For any value of $x$, $g(x)$ will always be greater than $h(x)$.
• $g(x)>h(x)$ for $x = - 1$.
• For positive values of $x$, $g(x)>h(x)$.
• For negative values of $x$, $g(x)>h(x)$.

Explanation:

Step1: Analyze the functions

$g(x)=x^{2}\geq0$ and $h(x)=-x^{2}\leq0$ for all real - valued $x$.

Step2: Check the first statement

Since $g(x)\geq0$ and $h(x)\leq0$ for all $x\in R$, for any value of $x$, $g(x)\geq0$ and $h(x)\leq0$, so $g(x)$ will always be greater than or equal to $h(x)$. When $x\neq0$, $g(x)>h(x)$. So the statement "For any value of $x$, $g(x)$ will always be greater than $h(x)$" is true.

Step3: Check the second statement

As $g(x)\geq0$ and $h(x)\leq0$ for all $x\in R$, $h(x)$ is not always greater than $g(x)$, so this statement is false.

Step4: Check the third statement

When $x = - 1$, $g(-1)=(-1)^{2}=1$ and $h(-1)=-(-1)^{2}=-1$. Since $1>-1$, $g(x)>h(x)$ for $x = - 1$, so this statement is true.

Step5: Check the fourth statement

When $x = 3$, $g(3)=3^{2}=9$ and $h(3)=-3^{2}=-9$. Since $9>-9$, $g(x)>h(x)$ for $x = 3$, so the statement $g(x)<h(x)$ for $x = 3$ is false.

Step6: Check the fifth statement

For positive values of $x$, let $x=a>0$. Then $g(a)=a^{2}>0$ and $h(a)=-a^{2}<0$. So $g(x)>h(x)$ for positive values of $x$, and this statement is true.

Step7: Check the sixth statement

For negative values of $x$, let $x=-a$ where $a>0$. Then $g(-a)=(-a)^{2}=a^{2}>0$ and $h(-a)=-(-a)^{2}=-a^{2}<0$. So $g(x)>h(x)$ for negative values of $x$, and this statement is true.