Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

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 $x = \frac{1}{2}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.