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

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Question

(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 $e^x = 1$,

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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 $x = 0$.Since the minimum value is $0$,the graph of $f(x)$ touches the $x$-axis only at $x = 0$.Therefore,the equation has exactly one real root,$x = 0$.

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

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