The smallest integer x for which the inequality $$\frac{x-7}{x^2 + 5x-36}$$ > 0 is given by
$$\frac{x-7}{x^2 + 5x-36}$$ > 0
$$\frac{x-7}{(x+9)(x-4)}$$ > 0
So $$x \epsilon (-9,4) U (7, \infty$$)
So smallest integer in this range is $$ x = -8$$.
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