Chapter 5.2, Problem 2AYU

Where is the function $f\left(x\right)=x^2$ increasing? Where is it decreasing? (p. 84)

To determine

To find: Where is the function $f\left(x\right)=x^2$ increasing? Where is it decreasing?

Answer to Problem 2AYU

Solution:

The left side of the $y\text{-axis}$ is decreasing and the right side of the $y\text{-axis}$ is increasing.

You can view this in this given graph.

i.e., the graph of $f\left(x\right)$ rises continually as we move from left to right and is called an increasing function. Also in the other side, the graph of $f\left(x\right)$ goes continually down as we move from left to right and is called a decreasing function.

Explanation of Solution

Given:

The function is $f\left(x\right)=x^2$.

Calculation:

By the definition,

A function is increasing on an interval if for any $x_1$ and $x_2$ in the interval then,

$x_1<x_2$ implies $f\left(x_1\right)<f\left(x_2\right)$

A function is decreasing on an interval if for any $x_1$ and $x_2$ in the interval then,

$x_1<x_2$ implies $f\left(x_1\right)>f\left(x_2\right)$

The given function is increasing on one interval and decreasing on another. The function $f\left(x\right)=x^2$ shown in the graph below is decreasing on the interval $\left(-\infty ,0\right]$ and increasing on $\left[0,\infty \right)$.

The function $f\left(x\right)=x^2$ is symmetrical about the $y\text{-axis}$, the $y\text{-axis}$ acts as a mirror reflecting the graph. The $y\text{-axis}$, $\left(x=0\right)$ is the axis of symmetry of the function. Functions that are symmetrical about the $y\text{-axis}$ are called even functions.

Also, $f\left(–1\right)={\left(–1\right)}^2=1=f\left(1\right)$. There are two values of $x$ that give the $y$ value 1 so the function is not one - to - one. $f\left(x\right)$ is a parabola and a horizontal line can cut it twice.