Q. 2C - Calculus

Question

Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) A function that is decreasing on $(-\infty ,0)$, increasing on $(0,\infty )$, and undefined at $x=0$.

(b) A function that is decreasing on $(-\infty ,0]$ and increasing on $[0,\infty )$.

(c) A function that is always positive and always decreasing, on all of $R$.

Step-by-Step Solution

Answer

(a). The example of the function that is decreasing on $(-\infty ,0)$, increasing on $(0,\infty )$, and undefined at $x=0$ is, $f\left(x\right)=\frac{x^6-1}{x^2}$.

(b). Therefore, the example of the function that is decreasing on $(-\infty ,0)$ increasing on $(0,\infty )$ is $f\left(x\right)=x^2$

(c). Therefore, the example of the function that is always positive and always decreasing, on all of $\mathrm{R}f\left(x\right)=e^{-x}$

Part(a) Step 1: Given information

A function that decreases on $(-\infty ,0)$, increases on $(0,\infty )$, and becomes undefined at $x=0$.

Part(a) Step 2: Explanation.

A purpose $f\left(x\right)$ is defined as an increasing function on the interval $\left(x_1,x_2\right)$, if  $x_1<( or >) x_2$  denotes that  $f\left(x_1\right)<( or >) f\left(x_2\right)$

A purpose $f\left(x\right)$ is defined as a decreasing function on an interval $\left(x_1,x_2\right)$ if  $x_1<( or >) x_2$ denotes that $f\left(x_1\right)>( or <) f\left(x_2\right)$.

Consider the function, $f\left(x\right)=\frac{x^6-1}{x^2}$.

At $x=0$, the above function is undefined.

Similarly, test that the function decreases on $(-\infty ,0)$ and increases on $(0,\infty )$ by changing the values in those intervals.

Therefore, the example of the function that is decreasing on $(-\infty ,0)$, increasing on $(0,\infty )$, and undefined at $x=0$ is, $f\left(x\right)=\frac{x^6-1}{x^2}$.

Part (b) Step 1: Given information.

A function that decreases on $(-\infty ,0]$ and increases on $[0,\infty )$

Part (b) Step 2: Explanation

A function $f\left(x\right)$ is defined as an increasing function on the interval $\left(x_1,x_2\right)$, if $x_1<( or >) x_2$ means that $f\left(x_1\right)<( or >) f\left(x_2\right)$

A function  $f\left(x\right)$ is said to be the decreasing function on an interval $\left(x_1,x_2\right)$, if $x_1<( or >) x_2$ implies that $f\left(x_1\right)>( or <) f\left(x_2\right)$

Consider the function $f\left(x\right)=x^2$

Check that the above function is decreasing on $(-\infty ,0)$ and increasing on $(0,\infty )$ by changing the values in those intervals.

Therefore, the example of the function that is decreasing on $(-\infty ,0)$ increasing on $(0,\infty )$ is $f\left(x\right)=x^2$

Part (c) Step 1: Given information.

A function that is always positive and always decreasing, on all of $R$.

Part (c) Step 2: Explanation

A purpose $f\left(x\right)$ is defined as an increasing function on the interval $\left(x_1,x_2\right)$ if  $x_1<( or >) x_2$ denotes that $f\left(x_1\right)<( or >) f\left(x_2\right)$

A function $f\left(x\right)$ is said to be a decreasing function on an interval $\left(x_1,x_2\right)$, if $x_1<( or >) x_2$ denotes that $f\left(x_1\right)>( or <) f\left(x_2\right)$

Consider the equation $f\left(x\right)=e^{-x}$.

Check that the above function is always positive and decreasing from any value of $x\in \mathrm{ℝ}$.

As a result, the function that is always positive and always decreasing, on all of $\mathrm{R}f\left(x\right)=e^{-x}$