over what interval is the function increasing, and over what interval is the function decreasing?\n|x|f(x)=8x…

over what interval is the function increasing, and over what interval is the function decreasing?\n|x|f(x)=8x²|(x,y)|\n|-2|32|(-2,32)|\n|-1|8|(-1,8)|\n|0|0|(0,0)|\n|1|8|(1,8)|\n|2|32|(2,32)|\nthe function f(x) is increasing over the interval . (simplify your answer. type an inequality.)

over what interval is the function increasing, and over what interval is the function decreasing?\n|x|f(x)=8x²|(x,y)|\n|-2|32|(-2,32)|\n|-1|8|(-1,8)|\n|0|0|(0,0)|\n|1|8|(1,8)|\n|2|32|(2,32)|\nthe function f(x) is increasing over the interval . (simplify your answer. type an inequality.)

Answer

Explanation:

Step1: Recall derivative rule

For $y = ax^n$, $y'=nax^{n - 1}$. For $f(x)=8x^{2}$, $f'(x)=16x$.

Step2: Find increasing - interval condition

A function is increasing when $f'(x)>0$. Set $16x>0$.

Step3: Solve the inequality

Dividing both sides of $16x>0$ by 16 gives $x > 0$.

Answer:

$x>0$