anemone said:Find the range of $k$ for which $\dfrac{e^x+e^{-x}}{2}\le e^{kx^2}$ holds for all real $x$.
chisigma said:[sp]Considering the Taylor expansions...
$\displaystyle \cosh x = 1 + \frac{x^{2}}{2} + ...\ (1)$
$\displaystyle e^{k\ x^{2}} = 1 + k\ x^{2} + ...\ (2)$
... the condition $\displaystyle \cosh x \le e^{k\ x^{2}}, \forall x$ is equivalent to say $\displaystyle k \ge \frac{1}{2}$...[/sp]
Kind regards
$\chi$ $\sigma$
The range of k for real x in the equation $\dfrac{e^x+e^{-x}}{2}\le e^{kx^2}$ is all real numbers greater than or equal to 0. This means that any value of k that makes the equation true for real values of x falls within this range.
To determine the range of k for a given equation, you can solve for k by isolating it on one side of the equation. In this case, you can manipulate the given equation to get k by itself on one side, and then determine the range of possible values for k based on the properties of the exponential function.
No, the range of k cannot be negative in this equation. This is because the exponential function has a minimum value of 0, meaning that the expression $e^{kx^2}$ can never be negative. Therefore, the range of k for this equation must be all real numbers greater than or equal to 0.
If k is equal to 0, the equation becomes $\dfrac{e^x+e^{-x}}{2}\le 1$, which is true for all real values of x. This is because any number raised to the power of 0 is equal to 1. Therefore, the range of k includes the value 0.
The range of k does not change if you change the values of x. This is because the range of k is dependent on the properties of the exponential function, not the values of x. As long as the values of x are real, the range of k will remain all real numbers greater than or equal to 0.