Chapter 1, Problem 11T

(a)

To determine

To evaluate: The solution of inequality $-4<5-3x\le 17$ in interval notation and sketch the solution on real number line.

Answer to Problem 11T

The solution of inequality $-4<5-3x\le 17$ is the interval $\left[-4,3\right)$.

Explanation of Solution

Given:

The inequality is $-4<5-3x\le 17$.

Calculation:

Section1:

Subtract same quantity from each side gives an equivalent inequalitythat is,

$A\le B\iff A-C\le B-C$

The solution of given inequality is all x-values that satisfy both the inequalities $-4<5-3x$ and $5-3x\le 17$ so use above rule of inequality and subtract 5 from both the inequality,

$\begin{array}{c}-4-5<-3x\le 17-5\\ -9<-3x\le 12\end{array}$

Multiply each side of inequality by negative quantity that is $\left(-1\right)$ which reverse the direction of inequality,

$\begin{array}{c}-12\le 3x<9\\ \frac{-12}{3}\le x<\frac{9}{3}\\ -4\le x<3\end{array}$

The set of solution consists all x-values from $x=-4$ to $x=3$ include the endpoint $-4$ because inequality is satisfy at the endpoint $-4$ so the interval notation of inequality is $\left[-4,3\right)$.

Thus,the solution of inequality $-4<5-3x\le 17$ is the interval $\left[-4,3\right)$.

Section2:

The solution of inequality from section 1 is $\left[-4,3\right)$ so graph of inequality is shown below,

Figure (1)

Figure (1) shows the solution of inequality which includes all x-values from $x=-4$ to $x=3$ and close interval at $-4$ and open interval at 3.

(b)

To determine

To evaluate: The solution of inequality $x\left(x-1\right)\left(x+2\right)>0$ in interval notation and sketch the solution on real number line.

Answer to Problem 11T

The solution of inequality $x\left(x-1\right)\left(x+2\right)>0$ is the interval $\left(-2,0\right)\cup \left(1,\infty \right)$.

Explanation of Solution

Given:

The inequality is $x\left(x-1\right)\left(x+2\right)>0$.

Calculation:

Section1:

The factors of the left-hand side are x, $\left(x-1\right)$, $\left(x+2\right)$ and these are zero when $x=-2,0,1$.

These values of x divide the real line into the intervals $\left(-\infty ,-2\right)$, $\left(-2,0\right)$, $\left(0,1\right)$ and $\left(1,\infty \right)$.

Now, make a table indicating the sign of each factor on each interval,

	$x<-2$	$-2<x<0$	$0<x<1$	$x>1$
x	$-$	$-$	$+$	$+$
$\left(x-1\right)$	$-$	$-$	$-$	$+$
$\left(x+2\right)$	$-$	$+$	$+$	$+$
$x\left(x-1\right)\left(x+2\right)$	$-$	$+$	$-$	$+$

From the above sign table, the inequality is satisfied on the intervals $\left(-2,0\right)$, $\left(1,\infty \right)$ and the end point of intervals does not satisfy the inequality.

Thus, the solution of inequality $x\left(x-1\right)\left(x+2\right)>0$ is the interval $\left(-2,0\right)\cup \left(1,\infty \right)$.

Section 2:

The given inequality is $x\left(x-1\right)\left(x+2\right)>0$.

The solution of inequality from section 1 is $\left(-2,0\right)\cup \left(1,\infty \right)$ so graph of inequality is shown below,

Figure (2)

Figure (2) shows the solution of inequality which includes all x-values from $x=-2$ to $x=0$ and all values after $x=1$ because these x-coordinates does not satisfy the inequality they shows open interval on real line number.

(c)

To determine

To evaluate: The solution of inequality $\left|x-4\right|<3$ in interval notation and sketch the solution on real number line.

Answer to Problem 11T

The solution of inequality $\left|x-4\right|<3$ is the interval $\left(1,7\right)$.

Explanation of Solution

Given:

The inequality is $\left|x-4\right|<3$.

Calculation:

Section1:

The inequality $\left|x-4\right|<3$ is equivalent to,

$-3<x-4<3$

The solution of above inequality is all x-values that satisfy both the inequalities $-3<x-4$ and $x-4<3$ so add 4 to each side of inequality to get solution,

$\begin{array}{c}-3+4<x<3+4\\ 1<x<7\end{array}$

Thus, the solution of inequality $\left|x-4\right|<3$ is the interval $\left(1,7\right)$.

Section2:

The solution of inequality from section 1 is $\left(1,7\right)$ so graph of inequality is shown below,

Figure (3)

Figure (3) shows the solution of inequality which includes all x-values from $x=1$ to $x=7$ and the endpoints does not satisfy the inequality so inequality shows open interval at both the sides on real number line.

(d)

To determine

To evaluate: The solution of inequality $\frac{2x-3}{x+1}\le 1$ in interval notation and sketch the solution on real number line.

Answer to Problem 11T

The solution of inequality $\frac{2x-3}{x+1}\le 1$ is the interval $\left(-1,4\right]$.

Explanation of Solution

Given:

The inequality is $\frac{2x-3}{x+1}\le 1$.

Calculation:

Section1:

First move all terms to the left-hand side of the inequality then factor the inequality to get values of x,

$\begin{array}{c}\frac{2x-3}{x+1}-1\le 0\\ \frac{2x-3-\left(x+1\right)}{x+1}\le 0\\ \frac{x-4}{x+1}\le 0\end{array}$

The factor of numerator is $\left(x-4\right)$ and denominator is $\left(x+1\right)$ and these are zero when $x=-1,4$.

These values of x divide the real line into the intervals $\left(-\infty ,-1\right)$, $\left(-1,4\right)$, and $\left(4,\infty \right)$.

Now, make a table indicating the sign of each factor on each interval,

	$x<-1$	$x=-1$	$-1<x<4$	$x=4$	$x>4$
$\left(x-4\right)$	$-$	$-$	$-$	0	$+$
$\left(x+1\right)$	$-$	0	$+$	$+$	$+$
$\frac{x-4}{x+1}$	$+$	undefined	$-$	0	$+$

From the sign table, the inequality is satisfied on the interval $\left(-1,4\right)$ and the end point $-1$ does not satisfy the inequality, endpoint 4 satisfy the inequality.

Thus, the solution of inequality $\frac{2x-3}{x+1}\le 1$ is the interval $\left(-1,4\right]$.

Section2:

The given inequality is $\frac{2x-3}{x+1}\le 1$.

The solution of inequality from section 1 is $\left(-1,4\right]$ so graph of inequality is shown below.

Figure (4)

Figure (4) shows the solution of inequality which includes all x-values from $x=-1$ to $x=4$ because these x-coordinates $-1$ does not satisfy the inequality they shows open interval and 4 satisfy the inequality they shows close interval on real number line.