By Dawkins P.
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Additional resources for Algebra/Trig Review (2006)(en)(98s)
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]. Do not get locked into the idea that the intervals will alternate as solutions as they did in the first problem. Also, notice that in this case the inequality was less than OR EQUAL TO, so we did include the endpoints in our solution.
So, for this graph I’ll change the range to −6π ≤ x ≤ 6π so we can get at least two traces of the curve showing. Here is the graph. Period = 6. aspx Algebra/Trig Review In the case of tangent we have to be careful when plugging x’s in since tangent sin x doesn’t exist wherever cosine is zero (remember that tan x = ). Tangent will not cos x exist at 5π 3π π π 3π 5π x = , − , − , − , , , , 2 2 2 2 2 2 and the graph will have asymptotes at these points. Here is the graph of tangent on 5π 5π the range − .
Aspx Algebra/Trig Review −4 ≤ 7 x − 10 ≤ 4 6 ≤ 7 x ≤ 14 6 ≤x≤2 7 In solving these make sure that you remember to add the 10 to BOTH sides of the inequality and divide BOTH sides by the 7. One of the more common mistakes here is to just add or divide one side. 5. 1 − 2 x < 7 Solution This one is identical to the previous problem with one small difference. −7 < 1 − 2 x < 7 −8 < −2 x < 6 4 > x > −3 Don’t forget that when multiplying or dividing an inequality by a negative number (-2 in this case) you’ve got to flip the direction of the inequality.