Function Inequality Proof

[turophile]

Here's the problem (from the section in my calculus text on the exponential):

By examining the graph of 1/t for t ? 1, show that for x > 0, x/(1+x) < log(1+x) < x.

My work:

y = 1/t = t^(–1)
y' = – t^(-2) ? 0 ? no minimum or maximum
y'' = 2t^(-3) ? 0 ? no inflection point; also y'' > 0 for t ? 1, so concave up (and therefore decreasing) for t ? 1

f(x) = x/(1+x)
g(x) = log(1+x)
h(x) = x

d/dx f(x) = d/dx 1 – 1/(1 + x) = d/dx 1 – (1 + x)^(–1) = 1/(1 + x)^2 > 0 for x > 0
d/dx g(x) = 1/(1+x) > f(x) for x > 0
d/dx h(x) = 1 > g(x) for x > 0

f(0) = 0/(1 + 0) = 0
g(0) = log(1 + 0) = 0
h(0) = 0

Since f(0) = g(0) = h(0), and since f'(x) < g'(x) < h'(x) for x > 0, therefore (by an earlier proof) we have f(x) < g(x) < h(x) for x > 0, QED.

My question:

I arrived at the proof's conclusion without reference to the behavior of the function 1/t for t ? 1. Also, since this came from a section on the exponential, I suspect I was supposed to use the exponential function somewhere in the proof. (All I did was differentiate the log function.) The textbook does not give answers to problems requiring proofs, so I'm not sure I carried out the proof with the approach it was looking for. Any guidance on how I could/should have done this proof differently?

 

[lookagain]

Here's the problem (from the section in my calculus text on the exponential):

By examining the graph of 1/t for t ? 1, show that for x > 0, x/(1+x) < log(1+x) < x.

\(\displaystyle The \ second \ inequality:\)

Claim the following to help with the 2nd inequality:

\(\displaystyle log(1 + x) < ln(1 + x)\)

-------------------------------------------------------------


\(\displaystyle log(1 + x) \ vs. \ \log_e{(1 + x)}, \ or\)

\(\displaystyle log(1 + x) \ vs. \ \frac{log(1 + x)}{log(e)} . \ . \ . \ . \ change \ of \ base \ rule\)

Because log(e) < 1, then

\(\displaystyle log(1 + x) < \log_e{(1 + x)}, \ or\)

\(\displaystyle log(1 + x) < ln(1 + x)\)


If it can be shown that ln(1 + x) < x, then it follows that log(x) < x.

Have e raised to each side:

\(\displaystyle e^{ln(1 + x)} \ \ vs. \ \ e^x\)


\(\displaystyle 1 + x \ \ vs. \ \ e^x\)

Because x > 0, and e^x = 1 + x + (all positive terms),

then, after subtracting (1 + x) from each side,

\(\displaystyle 0 \ < \ (all \ positive \ terms)\)


\(\displaystyle Then \ log(1 + x) \ < \ x, \ and \ the \ second \ inequality \ is \ established.\)

 

[turophile]

Forgive me, lookagain. I should have clarified that my textbook uses "log" to mean log_e. So I should have stated the problem using "ln(1+x)" instead of "log(1+x)". Can you help me connect an understanding of 1/t (t ? 1) to the proof?

 

[lookagain]

Here's the problem (from the section in my calculus text on the exponential):

By examining the graph of 1/t for t ? 1, show that for x > 0, x/(1+x) < log(1+x) < x.

My work:

y = 1/t = t^(–1)
y' = – t^(-2) ? 0 ? no minimum or maximum
y'' = 2t^(-3) ? 0 ? no inflection point; also y'' > 0 for t ? 1,

\(\displaystyle > \ > so \ concave \ up \ ( \ and \therefore \ decreasing) \ < \ < \\)

for t ? 1

Being concave up there does not mean it will be decreasing there. For example,
look at a parabola that opens up, and in the section to the right of the axis of
symmetry, it is increasing.

 

[turophile]

Yes, I should have said "strictly decreasing or stricty increasing" rather than "decreasing." To show that 1/t is decreasing, could I simply evaluate 1/t at two points on (t ? 1), say t = 2 and t = 3, and note that 1/2 > 1/3? Still, I don't see the connection between the graph of 1/t and what the problem requires to be shown regarding the inequality of the three functions f, g, and h.

 

[Deleted member 4993]

Here's the problem (from the section in my calculus text on the exponential):

By examining the graph of 1/t for t ? 1, show that for x > 0, x/(1+x) < log(1+x) < x.

My work:

y = 1/t = t^(–1)
y' = – t^(-2) ? 0 ? no minimum or maximum ? y'<0 for all 't' - hence y(t) is decreasing (as t increases) for all 't'

 

[turophile]

So my observations on 1/t should go something like this:

y = 1/t
dy/dt = -1/(t[sup:1k01a5ba]2[/sup:1k01a5ba])

Since dy/dt < 0 for t ? 0, 1/t is decreasing for t ? 0.

I'm still struggling with how to make that fact relevant to the proof of the inequality. (I think) I've proven the inequality without reference to the graph of 1/t, but presumably it's important to the proof or the problem wouldn't have mentioned it.