-x²+6x+7≥0 to interval notation. Created by kyeotnakyoet. Math

Solution set in Interval Notation: [-1, 7]Solution:1. Determine critical pöints:2. Determine three intervals:a. x < -1b. x > 7c. -1 < x < 73. Test each interval:4. Solution set in Interval Notation:[-1, 7]

-x²+6x+7≥0 to interval notation...

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kyeotnakyoet @kyeotnakyoet

April 2021 1 59 Report

-x²+6x+7≥0 to interval notation

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jomaeap

Solution set in Interval Notation:

[-1, 7]

Solution:

1. Determine critical pöints:

$- {x}^{2} + 6x + 7 \geqslant 0 \\ - 1 (- {x}^{2} + 6x + 7 \geqslant 0) - 1 \\ {x}^{2} - 6x - 7 \geqslant 0 \\ {x}^{2} - 6x - 7 = 0 \\ (x - 7)(x + 1) = 0 \\ x = 7 \: \: \: \: x = - 1$

2. Determine three intervals:

a. x < -1

b. x > 7

c. -1 < x < 7

3. Test each interval:

$1. for\: x < - 1 \: let \: x = - 2 \\ {x}^{2} - 6x - 7 \leqslant 0 \\ { - 2}^{2} - 6( - 2) - 7 \leqslant 0 \\ 4 + 12 - 7 \leqslant 0 \\ 9 \leqslant 0 \: false \\ 2. \: for \: x > 7 \: let \: x = 8 \\ {x}^{2} - 6x - 7 \leqslant 0 \\ { 8}^{2} - 6( 8) - 7 \leqslant 0 \\ 64 - 48 - 7 \leqslant 0 \\ 9 \leqslant 0 \: false \\ 3. \: for \: - 1 < x < 7 \: let \: x = 5 \\ {x}^{2} - 6x - 7 \leqslant 0 \\ { 5}^{2} - 6( 5) - 7 \leqslant 0 \\ 25 - 30 - 7 \leqslant 0 \\ - 12 \leqslant 0 \: true$

4. Solution set in Interval Notation:

[-1, 7]

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