The greatest value of the function f(x) = -1 | vedclass

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(D) Given the function $f(x) = -1 + \frac{2}{2^{x^2} + 1}$.To find the greatest value of $f(x)$,we need to maximize the term $\frac{2}{2^{x^2} + 1}$.$

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(D) Given the function $f(x) = -1 + \frac{2}{2^{x^2} + 1}$.To find the greatest value of $f(x)$,we need to maximize the term $\frac{2}{2^{x^2} + 1}$.$A$ fraction is maximized when its denominator is minimized.The denominator $2^{x^2} + 1$ is minimized when $2^{x^2}$ is minimized.Since $x^2 \ge 0$ for all real $x$,the minimum value of $x^2$ is $0$,which occurs at $x = 0$.At $x = 0$,$2^{x^2} = 2^0 = 1$.Thus,the minimum value of the denominator is $1 + 1 = 2$.The maximum value of the fraction is $\frac{2}{2} = 1$.Therefore,the greatest value of $f(x)$ is $f(0) = -1 + 1 = 0$.

The greatest value of the function $f(x) = -1 + \frac{2}{2^{x^2} + 1}$ is:

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