Question

QQ.Why does 
$e^x\geq x+1$
 ? Please explain clearly. Question for mods/stars only.​

Answer 1

Step-by-step explanation:

★ Concept :-

Here the concept of Integration and Relations have been used. We see that we are given equation to be proved. We have to it to be true. So for this we have many different methods. Here we shall firstly understand the method of relations and then we will use integration. Then we shall understand the method of using Relation.

Let's do it !!

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★ Solution :-

• Method I ::

We need to prove,

$\;\;\bf{\mapsto\;\;e^{x}\;\geq\;(x\:+\:1)}$

We already know that e^x is a continuous function. This means that it increases continuously with derivative. This eventually means that it's ending range is ∞.

Since e^x is continuously increasing this means that x ≥ 0 for all conditions. This definately means that e^x ≥ 1 as x is the exponent of e. Let's now use relation here. We know that, by Mean Value Theorem there exist an exponent y such that y ε (0, x) .

(Here ε represents 'belongs to').

The reason is that for the mean value of e^x, there will exists a function e^y where y belongs to (0, x).

Now this can be written as,

$\;\;\bf{\mapsto\;\;\blue{e^{x}\:-\:1\;=\;e^{y}\:\times\:x}}$

Now if we transpose 1 to the other side, the resultant expression will be,

$\;\;\bf{\mapsto\;\;\green{e^{x}\;=\;e^{y}\:\times\:x\;+\:1}}$

We already know that y ε (0, x) . This means that e^y depends upon e^x. So, the equation can be written as,

$\;\;\bf{\mapsto\;\;\purple{e^{x}\;\geq\;x\;+\;1}}$

Here we removed e^y and change the equality to inequality.

Hence proved.

• Method II ::

We already have proved above that e^x is an increasing continuous function. Now let's apply integration here with respect to t.

$\;\;\displaystyle{\bf{\rightarrow\;\;\orange{\int_{0} ^{x}(1\:-\:e^{t})dt\;\geq\;\int_{0} ^{x} 0 dt\;=\;0}}}$

Here ∀x ≥ 0 (since e^x is increasing)

$\;\;\displaystyle{\bf{\rightarrow\;\;x\;+\;e^{-x}\;-\;1\;=\;0}}$

$\;\;\displaystyle{\bf{\rightarrow\;\;x\;+\;e^{-x}\;=\;1}}$

Now we already know that ∀x ≥ 0, this means x will be positive.

$\;\;\displaystyle{\bf{\rightarrow\;\;x\;+\;e^{x}\;=\;1}}$

$\;\;\displaystyle{\bf{\rightarrow\;\;e^{x}\;=\;1\;-\;x}}$

Now we clearly see that x is coming to be -ve. This means we need to change the equality here to inequality.

On removing the negative sign, we get that e^x is greater or equal to L.H.S.

On changing the equality to inequality, we get

$\;\;\displaystyle{\bf{\rightarrow\;\;e^{x}\;\geq\;x\;+\;1}}$

Hence, proved.