1. ## mean value theorem

Use the mean value theorem to establish the following inequalities.

A) e^x > 1 + x for x>0
B) (x-1)/x < ln x < x-1 for x >1
e) sqrt(1+x) < 5 + (x-24)/10 for x>24

2. Originally Posted by luckyc1423
Use the mean value theorem to establish the following inequalities.

A) e^x > 1 + x for x>0
Consider f(t)=e^t on the interval [0,x]

Then, f is continous on [0,x] and differenciable on (0,x):

(f(x)-f(0))/(x-0) = f'(c) for c in (0,x)

Thus,
(e^x-1)/x = e^c
Thus,
e^x-1= x*e^c for c>0 but then, e^c >1
Thus,
e^x - 1 > x
Thus,
e^x > 1+x

3. Originally Posted by luckyc1423
B) (x-1)/x < ln x < x-1 for x >1
Consider f(t) = ln t on the interval [1,x].

Notice that f is continous on [1,x] and differenciable on (1,x).

By Lagrange's Mean Value Theorem:

(f(x)-f(1))/(x-1) = f'(c) for c in (1,x)

Thus,

ln(x)/(x-1) = 1/c for c in (1,x)

Notice that
1/x< 1/c < 1/1
Thus,
1/x < ln (x) / (x-1) < 1
Multiply by denominator,
(x-1)/x < ln x < (x-1)

4. Originally Posted by luckyc1423
e) sqrt(1+x) < 5 + (x-24)/10 for x>24
Consider,
f(t) = sqrt(1+t) on interval [24,x].

Thus, f is continous on [24,x] and differenciable on (24,x).

Use the Mean Value Theorem:

[sqrt(x+1) - sqrt(24+1)]/[x-24] = f'(c) = 1/2*sqrt(1+c) for c in (24,x).

Note that, 1/2*sqrt(1+c) < 1/2*sqrt(24+1) = 1/10

Thus,

[sqrt(x+1) - 5]/[x-24] < 1/10

Thus,

sqrt(1+x) < 5 + (x-24)/10

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# prove that 1 x<e^x by mean value theorem

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