x^2+6x-7/x^2+1 greater or equal to 2

Thursday, December 15, 2011

We are asked to solve the inequality $\displaystyle\frac{{{{x}}^{{2}}+{6}{x}-{7}}}{{{{x}}^{{2}}+{1}}}\ge{2}$

We can multiply both sides by x^2+1 since it is positive for all x:

$\displaystyle{{x}}^{{2}}+{6}{x}-{7}\ge{2}{\left({{x}}^{{2}}+{1}\right)}$

$\displaystyle{{x}}^{{2}}+{6}{x}-{7}\ge{2}{{x}}^{{2}}+{2}$

$\displaystyle{{x}}^{{2}}-{6}{x}+{9}\le{0}$

$\displaystyle{{\left({x}-{3}\right)}}^{{2}}\le{0}$

Since the square of a real number is nonnegative, this is true only at x=3.

The solution is x=3.

The graph:

Note that the graph approaches y=1 asymptotically"

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