First slide

Theory of equations

Question

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

Moderate

A

Only one real root $x = 0$
B

At least two real roots
C

Exactly two real roots
D

Infinitely many real roots

Solution

$e^{x} = x + 1 \Rightarrow 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + . . . . . = x + 1 \Rightarrow \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . = 0 x^{2} = 0, x^{3} = 0, . . . . x^{n} = 0$
Hence, $x = 0$ only one real root.

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