Question

Match the given expressions with the intervals in which their values could lie in

• A. 0,14
• B. [-7,∞)
• C. 3,∞
• D. $(0,\frac{1}{3}]$

Answer 1

$x^2-2x+5=x^2-2x+1+4$
$=(x-1{)}^2+4$

The minimum value of the expression is 4, when x = 1. The maximum value of the expression is $\infty$
Thus, $4\le (x-1{)}^2+4\le \infty$

Or, $\frac{1}{4}\ge \frac{1}{x^2-2x+5}>0$

Thus, the values of the expression $\frac{1}{x^2-2x+5}$ lie in the interval $(0,\frac{1}{4}]$

$x^2+4x-3=x^2+4x+4-7$
$=(x+2{)}^2-7$

The minimum value of the expression is -7, when x = -2. The maximum value of the expression is $\infty$
Thus, $-7\le (x+2{)}^2+-7\le \infty$

Thus, the values of the expression $x^2+4x-3$ lie in the interval $(-7,\infty$

$x^2-6x+18=x^2-6x+9+9$
$=(x-3{)}^2+9$

The minimum value of the expression is 9, when x = 3. The maximum value of the expression is $\infty$
Thus, $9\le (x-3{)}^2+9\le \infty$

and, $3\le \sqrt{(x-3{)}^2+9}\le \infty$

Or, $\frac{1}{3}\ge \frac{1}{\sqrt{x^2-6x+18}}>0$

Thus, the values of the expression $\frac{1}{\sqrt{x^2-6x+18}}$ lie in the interval $(0,\frac{1}{3}]$

$x^2-8x+25=x^2-8x+16+9$
$=(x-4{)}^2+9$

The minimum value of the expression is 9, when x = 4. The maximum value of the expression is $\infty$
Thus, $9\le (x-4{)}^2+9\le \infty$

Or, $3\le \sqrt{x^2-8x+25}\le \infty$


Thus, the values of the expression $\sqrt{x^2-8x+25}$ lie in the interval $(3,\infty )$