#### To determine

**To find:**

Positive number *a* that satisfies the given inequality

#### Answer

a=e

**Given: **An inequality ax≥1+x.

**Formulae used:**

lnax=xlna

#### Explanation

Consider ax≥1+x

Note that ax is defined only for positive values of *a.* Besides for all values of *a*, the value of ax is positive. But 1+x≤0 for x≤−1 . So the inequality is true for all *a* if x≤−1.

Thus we may evaluate the truthness of the inequality only for x∈(−1,∞)

Taking logarithm on both sides of equation, we have

xlna≥ln(1+x)

For *x=0* this inequality is satisfied by all values of *a*.

When x∈(−1,0)

we have

lna≤ln(1+x)x (∵division by a negative number reverses an inequality)

and when x∈(0,∞) we have

lna≥ln(1+x)x

So the positive number(s) *a * satisfying ax≥1+x is such that

lna≤ln(1+x)x for -1<x<0andlna≥ln(1+x)x for 0<x<∞

So to find bounds of *a* we need to find minimum of f(x)=ln(1+x)x in (−1,0) and maximum of f(x)=ln(1+x)x in (0,∞).

Let us differentiate *f* using quotient rule

f'(x)=11+xx−ln(1+x)x2

We may note that the function has a singularity at *x=0. *But we are not interested in it.

Since it is sum, difference, product and quotient of continuous functions f'(x)=11+xx−ln(1+x)x2=x−(1+x)ln(1+x)(1+x)x2

is also a continuous function.

Now f'(e−1)=e−1−(1+e−1)log(1+e−1)(1+e−1)(e−1)2=e−1−(e)log(e)(e)(e−1)2=e−1−e(e)(e−1)2=−1e(e−1)2<0

Similarly

f'(−1+1e)=(−1+1e)−(1+(−1+1e))log(1+(−1+1e))(1+(−1+1e))(−1+1e)2=(−1+1e)−(1e)log(1e)(1e)(1e−1)2=(−1+1e)+(1e)(1e)(1e−1)2=−1+2e(1e)(1e−1)2

But e>2⇒1>2e⇒0>2e

Thus

f'(−1+1e)=−1+2e(1e)(1e−1)2<0

When *f’(x)=0* we have

f'(x)=x−(1+x)ln(1+x)(1+x)x2=0⇒x−(1+x)ln(1+x)=0⇒x1+x=ln(1+x)⇒ex1+x=1+x

Since exponential is a one-one function this has only one solution. Clearly this solution is 0. Thus in both the intervals (−1,0) and (0,∞) we have that *f’(x)* is continuous, an also that it never takes value zero. Moreover in both intervals *f’(x)* takes negative value, for instance consider e−1 and −1+1e . Thus we may conclude by intermediate value theorem that *f’(x)* is always negative in both the intervals. (if it is to take a positive value in either of the interval then by IVT it will have to take zero somewhere, which is not true, so).

Thus we have that the function is decreasing in both the intervals (−1,0) and (0,∞)

So the minimum in (−1,0) and the maximum in (0,∞) is both given by the limit of *f* as it approaches 0.

So it follows that in (−1,0)f(x)=log(1+x)x≥limx→0f(x)=limx→0log(1+x)x

And in (0,∞).

f(x)=log(1+x)x≤limx→0f(x)=limx→0log(1+x)x

Note that limx→0log(1+x)x is in 0/0 form. So we shall use L’ hospital rule to evaluate it.

Thus

limx→0log(1+x)x=limx→011+x1=limx→011+x=1

Thus

log(1+x)x≥1 for -1<x<0⇒loga≤1 (∵loga≤log(1+x)x) andlog(1+x)x ≤1 for 0<x<∞⇒loga≥1 (∵loga≥log(1+x)x)

We may see this from the graph of *f(x)* as well,

From the graph also we can see that

log(1+x)x≥1 for -1<x<0⇒loga≤1 (∵loga≤log(1+x)x) andlog(1+x)x ≤1 for 0<x<∞⇒loga≥1 (∵loga≥log(1+x)x)

Combining both inequalities we have

loga=1⇒a=e

Thus *a=e* is the only number such that ax≥1+x for all *x. *One may verify this graphically using any graphing calculator, by plotting the graphs for various values of *a*.