Solve Inequality: x^2-4|-3|<1 | UofT Calculus

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To solve the inequality ||(x^2)-4|-3|< 1, it's effective to graph the function f(x) = ||x^2 - 4|-3| for better insight. A systematic approach involves dividing the problem into cases based on the value of x: for x^2 ≤ 4, the function simplifies to |x^2 - 1| < 1, while for x^2 > 4, it becomes |x^2 - 7| < 1. Each case leads to specific conditions for x that need to be analyzed further. Additionally, it's important to show your attempts at solving the equation when seeking help in discussions. Understanding these steps is crucial for tackling similar inequalities in calculus.
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Hey everyone. I'm taking Calculus at UofT and I got a question in a problem set that kind of got me thinking, and well, I'm not sure if I'm doing it correctly. This isn't the exact question, but, how would you go about solving this inequality:
||(x^2)-4|-3|< 1
 
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AnthonyJohnAre said:
Hey everyone. I'm taking Calculus at UofT and I got a question in a problem set that kind of got me thinking, and well, I'm not sure if I'm doing it correctly. This isn't the exact question, but, how would you go about solving this inequality:
||(x^2)-4|-3|< 1

The easiest way is to first gain some insight into the problem by plotting a graph of the function f(x) =||x^2 - 4|-3|.

A systematic way is to split up x into cases: (1) x^2 ≤ 4; (2) x^2 > 4. In case (1) we have f(x) = |4-x^2-3| = |1-x^2| = |x^2-1|. So, for x^2 ≤ 4 we also want |x^2-1| < 1. What do these requirements say about x? In case (2) we have f(x) = |x^2 - 4 - 3| = |x^2-7|, so for x^2 > 4 we also need |x^2-7| < 1. What do these tell you about x?
 
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AnthonyJohnAre said:
Hey everyone. I'm taking Calculus at UofT and I got a question in a problem set that kind of got me thinking, and well, I'm not sure if I'm doing it correctly. This isn't the exact question, but, how would you go about solving this inequality:
||(x^2)-4|-3|< 1

Please check your PMs. You should show your attempt at solving the equation as part of your first post asking for help.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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