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Sagot :
Answer:
For the inequality to be true, both factors must have the same sign or one factor must be 0. The zeroes occur at x = − 2 , x = 5 . Thus, our intervals will be ( − ∞ , − 2 ] , [ − 2 , 5 ] , [ 5 , ∞ ) Take a sample point from each. A sample point of − 3 for the first interval yields ( − 3 + 2 ) ( − 3 − 5 ) = ( − 1 ) ( − 8 ) = 8 , so the first interval satisfies the inequality. Choosing a sample point of x = 0 for the second interval, we receive ( 2 ) ( − 5 ) = − 10 , so this interval will not satisfy the equality. A sample point of x = 6 for the third interval gives us ( 8 ) ( 1 ) = 8 , so the third interval satisfies. Thus, the domain that satisfies the inequality is ( ∞ , − 2 ] & [ 5 , ∞ ) On the number line , you would draw and fill in a circle above -2, and draw an arrow from that point to the left. You would then draw and fill in a circle above 5, and provide a similar arrow to the right (i.e. towards positive infinity). We fill in these circles because the intervals are closed at these points, i.e. these points satisfy the inequality. If we instead had ( x + 2 ) ( x − 5 ) > 0 , the circles would be empty because those points would not satisfy.
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