Chapter 1, Problem 13T

To determine

To find: The values of x for the expression $\sqrt{6x-x^2}$, where x is defined as real number.

Answer to Problem 13T

The values of x in $0\le x\le 6$ defined as a real numbers.

Explanation of Solution

The expression is $\sqrt{6x-x^2}$.

The expression $\sqrt{6x-x^2}$ is defined as a real number when polynomial function $6x-x^2$ is greater than or equal to zero.

The square roots of any function is not defined for negative values of function.

Thus,

$6x-x^2\ge 0$

On further simplification,

$x\left(6-x\right)\ge 0$

The factors of the left-hand side are x, $\left(6-x\right)$ and these are zero when $x=0,6$.

Two possible values of x give the corresponding intervals of inequality $\left(-\infty ,0\right)$, $\left(0,6\right)$ and $\left(6,\infty \right)$.

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

	$x<0$	$x=0$	$0<x<6$	$x=6$	$x>6$
x	$-$	0	$+$	$+$	$+$
$\left(6-x\right)$	$+$	$+$	$+$	0	$-$
$x\left(6-x\right)$	$-$	0	$+$	0	$-$

Table (1)

From the above sign table, the inequality is satisfied on the intervals $\left(0,6\right)$ and the end points of intervals $x=0$ and $x=6$.

So the solution of given expression is all x-values, in between $x=0$ to $x=6$, for which the expression $\sqrt{6x-x^2}$ is defined as a real number.

Thus, the values of x in $0\le x\le 6$ defined as a real numbers.