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Is there any way to solve the inequation: $${e^{2x - 1}} \ge 2x$$ without plotting the graph?

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up vote 4 down vote accepted

All you need to know is $e^x \geq 1 + x$, $\forall x \in \mathbb{R}$

Equality holds when $x=0$

To prove $e^x \geq 1 + x$, $\forall x \in \mathbb{R}$, consider $f(x) = e^x - 1 -x$

$f'(x) = e^x - 1 = 0 \Rightarrow$, $x=0$ is an extremum.

Further $f''(x) = e^x > 0$, $\forall x \in \mathbb{R}$

Hence, $x=0$ is a minimum

Hence, we get $f(x) \geq f(0)$, $\forall x \in \mathbb{R}$

Hence, $e^x - 1 - x \geq 0$, $\forall x \in \mathbb{R}$

Hence, $e^x \geq 1 + x$, $\forall x \in \mathbb{R}$

Replacing $x$ by $2x-1$, we get $e^{2x-1} \geq 1 + (2x-1) = 2x$, $\forall x \in \mathbb{R}$

Equality holds when $2x-1=0$ i.e. when $x = \frac{1}{2}$

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