Chapter 2.4, Problem 28AYU

If $f\left(x\right)=\{\begin{array}{l}{x}^3\text{if}-2\le x<1\\ 3x+2\text{if}1\le x\le 4\end{array}$

find: (a) $f\left(-1\right)$

(b) $f\left(0\right)$

(c) $f\left(1\right)$

(d) $f\left(3\right)$

To determine

To find: The values of $f\left(-1\right),f\left(0\right),f\left(1\right)$and$f\left(3\right)$ using the given piecewise defined function $f\left(x\right)=\{\begin{array}{c}{x}^3 if-2\le x<1\\ 3x+2 if 1\le x\le 4\end{array}$

Answer to Problem 28AYU

Solution:

The values of $f\left(-1\right),f\left(0\right),f\left(1\right)$and$f\left(3\right)$using the given piecewise defined function are $-1,0,5\text{ and }11$.

Explanation of Solution

Given:

The function is: $f\left(x\right)=\{\begin{array}{c}{x}^3 if-2\le x<1\\ 3x+2 if 1\le x\le 4\end{array}$

It is asked to find the values of $f\left(-1\right),f\left(0\right),f\left(1\right)$ and $f\left(3\right)$ using the piecewise defined function.

1. Consider $f\left(-1\right),-1$ belongs to the condition $-2\le x<1$.

$f\left(x\right)=x^3$

$f\left(-1\right)={\left(-1\right)}^3=-1$

2. Consider $f\left(0\right),0$ belongs to the condition $-1\le x<2$.

$f\left(x\right)=x^3$

$f\left(0\right)={\left(0\right)}^3=0$

3. Consider $f\left(1\right),1$ belongs to the condition $1\le x\le 4$.

$f\left(x\right)=3x+2$

$f\left(1\right)=3\left(1\right)+2=5$

4. Consider $f\left(3\right),3$ belongs to the condition $1\le x\le 4$.

$f\left(x\right)=3x+2$

$f\left(3\right)=3\left(3\right)+2=11$

The values of $f\left(-1\right),f\left(0\right),f\left(1\right)$ and $f\left(3\right)$ using the given piecewise defined function are $-1$, 0, 5 and 11.