MHB What is the Range of x for $|x|^{x^2-3x-4}<1$?

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The inequality $|x|^{x^2-3x-4}<1$ requires analyzing the expression to determine the range of values for $x$. The critical points occur when $x^2 - 3x - 4 = 0$, leading to roots at $x = 4$ and $x = -1$. The solution reveals that the inequality holds for $x$ in the intervals $(-\infty, -1)$ and $(4, \infty)$. Evaluating the range $(a, b)$, where $a = -1$ and $b = 4$, results in $b - a = 5$. The final answer is that the difference $b - a$ equals 5.
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If the range of values of $x$ that satisfies $|x|^{x^2-3x-4}<1$ is given by $(a,\,b)$, evaluate $b-a$.

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Congratulations to kaliprasad for his correct solution, which is shown below:):

$|x|^{(x-4)(x+1)} \lt 1$

$(x-4)(x+1)> 0$ for $x < -1$ or $x > 4$

$(x-4)(x+1)<0$ for $x > -1$ and $x< 4$

$(x-4)(x+1)=0$ at $x=-1$ or $x=4$

It is true if either

1) $|x| < 1$ and $(x-4)(x+1) > 0$ possible

or

2) $|x| > 1$ and $(x-4)(x+1) < 0$ So we get $a = 1$ and $b = 4$ therefore $b-a = 3$.
 
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