Chapter 4, Problem 41RE

To determine

To solve: The inequality $x^3+x^2<4x+4$ . Also, graph the solution set.

Answer to Problem 41RE

The interval is $\left[-2,-1\right]\cup [2,\infty )$ .

Explanation of Solution

Given information:

The inequality $x^3+x^2<4x+4$ .

Formula used:

The zeros of the function $f(x)=0$ .

Calculation:

Consider ,

  $x^3+x^2<4x+4$

Step 1.

Re-arranging the inequality

  $=x^3+x^2-4x-4<0$ .

Step 2.

The zeros of $f(x)=x^3+x^2-4x-4$ .

Using , hit and trial method

Put $x=-1$ we get,

  $\begin{array}{l}={(}^{-}+{(}^{-}-4(-1)-4\\ =-1+1+4-4\\ =0\end{array}$

Using long division method we have,

  $x^2$

  $x+1\overline{)\begin{array}{l}{x}^3+x^2-4x-4\\ x^3+x^2\end{array}}$

  $x+1\overline{)\begin{array}{l}{x}^3+x^2-4x-4\\ -x^3-x^2\end{array}}$

  $x+1\overline{)-4x-4}$

  $-4$

  $x+1\overline{)\begin{array}{l}-4x-4\\ -4x-4\end{array}}$

  $x+1\overline{)\begin{array}{l}-4x-4\\ 4x+4\end{array}}$

The quotient is $x^2-4$ and remainder is 0.

Now,

  $=x^3+x^2-4x-4=(x+1)(x^2-4)$

  $=(x+1)(x-2)(x+2)=0$

  $\begin{array}{l}=x+1=0\\ =x=-1\end{array}$ or $\begin{array}{l}=x-2=0\\ =x=2\end{array}$ or $\begin{array}{l}=x+2=0\\ =x=-2\end{array}$

Step 3.

Using, the zeros to separate the real number line into one interval:

  $(-\infty ,-2);(-2,-1);(-1,2):(2,\infty )$

Step 4.

Evaluate the $f(x)=x^3+x^2-4x-4$ at each number.

Interval	$\left(-\infty ,-2\right)$	$\left(-2,-1\right)$	$\left(-1,2\right)$	$\left(2,\infty \right)$
Number chosen	$-3$	$-1.5$	1	3
Value of   $f$	$\begin{array}{l}={(}^{-}+{(}^{-}-4(-3)-4\\ =-27+9+12-4\\ =-10\end{array}$	$\begin{array}{l}={(}^{-}+{(}^{-}-4(-1.5)-4\\ =3.375+2.25+6-4\\ =7.625\end{array}$	$\begin{array}{l}={(}^1+{(}^1-4(1)-4\\ =1+1-4-4\\ =-6\end{array}$	$\begin{array}{l}={(}^3+{(}^3-4(3)-4\\ =27+9-12-4\\ =20\end{array}$
Conclusion	Negative	Positive	Negative	Positive

Now, using the above values we can draw a number line as follows:

  

Figure 1.

Step 5.

Using the information gathered from Step 1 − Step 4 we concluded that $f(x)\ge 0$ for all numbers x for $-2\le x\le -1$ and $x\ge 2$ . Hence, the solution set of the inequality $x^3+x^2<4x+4$ is $\left[-2,-1\right]\cup [2,\infty )$ .