Inequalities problem 1 by Dhaval Furia

If we take the derivative of $f(x) = 2^x + 2^{-x}$ and set it equal to zero, we obtain:

$f'(x) = 2^x \ln(2) + (\frac{1}{2})^x \ln(\frac{1}{2}) = [2^x - (\frac{1}{2})^x] \cdot \ln(2) = 0 \Rightarrow x = 0$

as the critical point. Evaluating the second derivative at this same critical point yields:

$f''(x) = [2^x + (\frac{1}{2})^x][\ln(2)]^2 \Rightarrow f''(0) = 2\ln^{2}(2) > 0$

hence, the global minimum for $f(x)$ occurs at the point $(0,2)$ . The function $g(x) = 2\cos(x^2 + x)$ has a maximum value of 2, which is attained at this same critical point $\Rightarrow$ only one real root exists.