Answer to Question #181806 in Calculus for desmond

Question

Use the Mean Value Theorem to establish the following inequalities (a) x < sin−1 x < x √ 1 − x 2 for 0 < x < 1. (b) e a (x − a) < ex − e a < ex (x − a) if a < x. (c) e x > x + 1, ∀ x > 0. (d) |sin x − sin y| ≤ |x − y| ∀x, y ∈ R. (e) x 1 + x < ln(1 + x) < x for −1 < x < 0 , x > 0 

Answer 1

a) x < sin^-1 x < x√ 1-x² ∀ x ∈ (0,1)

Let f(x) = sin^-1 x

f'(x) = 1/ √ 1-x²

From Mean Value Theorem,

[f(b) - f(a)] / (b-a) = f'(c)

Let 0 < c < x < 1

So from MVT we have

(sin^-1 x - sin^-1 0) / x = 1/ √ 1-c²

sin^-1 x / x = 1/ √ 1-c² ................................(1)

0 < c < x

0 < c² < x²

1-x² < 1-c² < 1

1 < 1/ √ 1-c² < 1/ √ 1-x²

x < sin^-1 x/ x < x/ √ 1-x² .................from equation (1)

b) e ^a (x-a) < e^x - e ^a < e ^x (x-a) if a<x

Let f(x) = e^x

f'(x) = e^x

From Mean Value Theorem,

f(b) - f(a)/ (b-a) = f'(c) ∀ c ∈ (a,b)

let a < x < b

so

e^x - e ^a /(x-a) = e ^c .............................(2)

a < c < x

e ^a < e ^c < e^x

e ^a (x-a) < e^x - e ^a < e ^x (x-a) ..................from equation (2)

c) e^x > x+1 ∀ x >0

Let f(x) = e^x - x- 1

f'(x) = e^x -1

Now since x> 0

So f'(x) >0

Hence, f(x) is an increasing function ∀ x > 0

So e^x -x-1>0

e^x > x+1

d) |sin x - sin y| <= |x - y| ∀ x,y ∈ R

Let f(x) = sin x

f'(x) = cos x

From Mean Value Theorem

f(b) - f(a) / (b-a) = f'(c)

sin b - sin a /( b-a) = cos c

| (sin b - sin a ) / (b-a) | = | cos c | .............(1)

Now, | cos c | <= 1 ...............(2)

| (sin b - sin a ) / (b-a) | <=1 .............from equation (1) and (2)

| (sin b - sin a ) | <= | b-a|

Replacing b → x and a → y we have

| (sin x - sin y ) | <= | x-y |

e) x/ (1+x) < ln (1+x) < x ∀ x ∈ (-1,0)

let f(x) = ln x

f'(x) = 1/x

From Mean Value Theorem

f(b) - f(a) / (b-a) = f'(c)

ln (b) - ln (a) = (b-a) /c .....................(1)

a < c < b

1/b < 1/c < 1/a

(b-a)/b < (b-a)/c < (b-a)/a

(b-a)/b < ln(b) - ln(a) < (b-a)/a

Replacing b → 1+x and a → 1 we have

x/(1+x) < ln(1+x) < x