Correction: That is not a logarithmic function, but here is how to find the critical numbers for the function. Explanation: \displaystyle{f{{\left({x}\right)}}}={e}^{{x}}-{2}{e}^{{-{x}}}-{3}{x} ...

You can use the theorem : p( a prime) divides the product ab if and only if p divides a or p divides b. Now x^{2} \equiv y^{2} mod p if and only if p divides x^{2}-y^{2}=(x-y)(x+y), ...

Jim H May 23, 2015 I assume that you want to find \displaystyle\frac{{d}}{{\left.{d}{x}\right.}}{\left({e}^{{{x}+{y}}}+{e}^{{{2}{x}}}+{e}^{{{2}{y}}}\right)} . If I have misunderstood ...

Frederico Guizini S. Apr 10, 2017 See the answer below:

Some basic manipulation of the quadratic in y in this expression gives the following inequality: e^{2x}(x+y^2+2y)\ge e^{2x}(x-1) then differentiation yields 2e^{2x}\left(x-\frac{1}{2}\right) = 0 \Rightarrow x_{min} = \frac{1}{2} ...

They are equivalent, in as much as \int_{\boldsymbol \Omega} (x_1 x_2 + y_1 y_2)^2 f_{\boldsymbol X}(\boldsymbol x) \, d \boldsymbol x = \int_{\boldsymbol \Omega} (x_1 x_2)^2 f_{\boldsymbol X}(\boldsymbol x) \, d \boldsymbol x + \int_{\boldsymbol \Omega} (y_1 y_2)^2 f_{\boldsymbol X}(\boldsymbol x) \, d \boldsymbol x + 2\int_{\boldsymbol \Omega} (x_1 x_2 y_1 y_2) f_{\boldsymbol X}(\boldsymbol x) \, d \boldsymbol x. ...