Number of integral values of x satisfying the inequality

What is the number of integral values of $$x$$ satisfying the inequality:

$$\frac{(e^x-1)(\sin(x)-2)(x^2-5x+4)}{x^2(-x^2+x-2)(2x+3)}\le 0$$

I was able to find three solutions: $$0$$, $$1$$ and $$4$$. Is there any other solution?

• Is it correct now? – Dr. Sonnhard Graubner May 20 at 16:58
• Yes , thanks sir – Abhishek Kumar May 20 at 16:58
• Function is not defined at $0$ – Andrei May 20 at 17:09
• But 0/0 is zero I think it is defined – Abhishek Kumar May 20 at 17:13
• Why $0$? Try plotting just $(e^x-1)/x^2$ The function has a discontinuity at $0$. On one side it goes to $+\infty$, on the other side it goes to $-\infty$ – Andrei May 20 at 17:21

Note the $$x^2$$ in the denominator ensures the function is not defined at $$x=0$$, otherwise has no effect on the inequality, so we may ignore it. Similarly, $$\sin x -2$$ is always negative, and so is $$-x^2+x-2$$, so both may be together ignored. As $$e^x-1$$ has the same sign as $$x$$, essentially we can substitute that, and equivalently solve for $$\frac{x(x-1)(x-4)}{2x+3} \leqslant 0$$
The intervals to check are $$x< -\frac32,x \in (-\frac32, 0), x \in (0, 1), x \in [1, 4]$$ and $$x> 4$$, which is easily done to get $$x \in (-\frac32,0) \cup [1, 4]$$, so integral solutions are $$x\in \{-1, 1, 2, 3, 4\}$$.