Dawkins P.'s Algebra/Trig Review (2006)(en)(98s) PDF

By Dawkins P.

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Example text

The points found above will divide the number line into regions in which the inequality will either always be true or always be false. So, pick test points from each region, test them in the inequality and get the solution from the results. For this problem the numerator will be zero at x = 3 and the denominator will be zero at x = −2 . The number line, along with the tests is shown below. So, from this number line it looks like the two outer regions will satisfy the inequality. We need to be careful with the endpoints however.

Tan  −  and tan    4  4  Solution 7π π π 7π Here we should note that = 2π − so and − are in fact the same angle! aspx Algebra/Trig Review Now, if we remember that tan ( x ) = sin ( x ) we can use the unit circle to find the cos ( x ) values the tangent function. So,  7π   π  sin ( − π 4 ) − 2 2 = = −1 . tan  tan  −  = = 2 2  4   4  cos ( − π 4 ) π  On a side note, notice that tan   = 1 and we se can see that the tangent function is 4 also called an odd function and so for ANY angle we will have tan ( −θ ) = − tan (θ ) .

Since the inequality was a strict inequality, we don’t include the endpoints since these are the points that make both sides of the inequality equal! 2. x 4 + 4 x 3 − 12 x 2 ≤ 0 Solution We’ll do the same with this problem as the last problem. x 4 + 4 x 3 − 12 x 2 ≤ 0 ⇒ x 2 ( x 2 + 4 x − 12 ) ≤ 0 ⇒ x 2 ( x + 6 )( x − 2 ) ≤ 0 In this case after factoring we can see that the left side will be zero at x = −6 , x = 0 and x = 2 . From this number line the solution to the inequality is −6 ≤ x ≤ 2 or [-6,2].

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Algebra/Trig Review (2006)(en)(98s) by Dawkins P.


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