Q. 72 - Calculus

Question

Find the intervals on which each function in Exercises $67-74$ is positive or negative. Make clear how your work uses the Intermediate Value Theorem and continuity. You may assume that polynomials and their quotients are continuous on the intervals on which they are defined.

$f\left(x\right)=\left\{\begin{array}{l}3x+1, if x<0\\ x, if x\ge 0\end{array}\right.$

Step-by-Step Solution

Answer

The function is positive on $\left(-\frac{1}{3},\infty \right)$ and negative on  $\left(-\infty ,-\frac{1}{3}\right)$.  

Step 1. Given Information

We are given the function $f\left(x\right)=\left\{\begin{array}{l}3x+1, if x<0\\ x, if x\ge 0\end{array}\right.$ and we need to find the intervals in which the function is positive or negative using the Intermediate Value Theorem and continuity.

Step 2. Finding the intervals

The piecewise-defined function f can be discontinuous only at its break point $x=0$.

Furthermore, its first component $3x+1$ is zero only at $x=-\frac{1}{3}$, its second component $x$ is zero only at $x=0$.The function f can change sign only at the roots and discontinuities at $-\frac{1}{3},0$.

Step 3. Testing the sign of f ( x )

Testing the sign of $f\left(x\right)$ to find the interval where the function is positive or negative:

When $x=-2$,

$$\begin{gathered} f(-2) \\ =3(-2)+1 \\ =-6+1 \\ =-5<0 \end{gathered}$$

When $x=-\frac{1}{5}$,

$$\begin{gathered} f(-\frac{1}{5}) \\ =3(-\frac{1}{5})+1 \\ =-\frac{3}{5}+1 \\ =\frac{2}{5}>0 \end{gathered}$$

When $x=2$,

$$\begin{gathered} f\left(2\right) \\ =2>0 \end{gathered}$$

Therefore, the function is positive on $\left(-\frac{1}{3},\infty \right)$ and negative on $\left(-\infty ,-\frac{1}{3}\right)$.