The equation e^x - x - 1 = 0 has | vedclass

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(A) Let $f(x) = e^x - x - 1$. To find the roots,we analyze the function $f(x)$. The derivative is $f'(x) = e^x - 1$. Setting $f'(x) = 0$ gives $e^x =

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Only one real root $x = 0$

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(A) Let $f(x) = e^x - x - 1$. To find the roots,we analyze the function $f(x)$. The derivative is $f'(x) = e^x - 1$. Setting $f'(x) = 0$ gives $e^x = 1$,which implies $x = 0$. At $x = 0$,$f(0) = e^0 - 0 - 1 = 1 - 1 = 0$. Since $f'(x)  0$ for $x > 0$,the function $f(x)$ has a global minimum at $x = 0$ with $f(0) = 0$. Therefore,$f(x) \ge 0$ for all real $x$,and $f(x) = 0$ only at $x = 0$. Thus,the equation has exactly one real root,$x = 0$.

The equation $e^x - x - 1 = 0$ has

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