pls help!!!!! due soon

To find all values of $x$ for which $f(x) > g(x)$, where $f(x) = |x|$ and $g(x) = x^2$, we need to compare their values for different ranges of $x$.

 

1. For $x \geq 0$:

$$f(x) = |x| = x$$

$$g(x) = x^2$$

 

So, $f(x) > g(x)$ for $x \geq 0$ when $x^2 > x$, which is true when $x > 1$ or $x < 0$.

 

2. For $x < 0$:

$$f(x) = |x| = -x$$

$$g(x) = x^2$$

 

So, $f(x) > g(x)$ for $x < 0$ when $-x > x^2$, which is true when $-x > x^2$ for $x < 0$.

 

Let's solve $-x > x^2$ for $x < 0$:

$$-x > x^2$$

$$x^2 + x < 0$$

$$x(x + 1) < 0$$

 

This inequality holds true when either $x < 0$ and $x + 1 > 0$, or $x > 0$ and $x + 1 < 0$.

 

1. For $x < 0$, $x(x + 1) < 0$ when $x < 0$ and $x + 1 > 0$ (the product of two negative numbers is positive):

$$x < 0$$

$$x + 1 > 0$$

$$x > -1$$

 

So, for $x < 0$, $f(x) > g(x)$ when $-1 < x < 0$.

 

Combining both ranges of $x$, we have $f(x) > g(x)$ for $x > 1$ or $-1 < x < 0$.

 

Therefore, all values of $x$ for which $f(x) > g(x)$ are $x \in (-1, 0) \cup (1, \infty)$.