Question

Use the derivative to show why f(x)=the integral of 1/square root of (4+e^t) from 0 to ln(x) has an inverse function?

Answer 1

prove that f(x) is strictly increasing/decreasing by showing that the derivative is always positive or negative

Answer 2

I think that f(x) would be increasing, but how would I prove it?

Answer 3

\[f(x) = \int \limits_0^{\ln x} \dfrac{1}{\sqrt{4+e^t}} dt\]

\[\implies f'(x) = \dfrac{d}{dx}\int \limits_0^{\ln x} \dfrac{1}{\sqrt{4+e^t}} dt = ? \]

Answer 4

use FTC + chainrule

Answer 5

\[f'(x) = \dfrac{d}{dx}\int \limits_0^{\ln x} \dfrac{1}{\sqrt{4+e^t}} dt = \dfrac{1}{\sqrt{4+e^{\ln x}}} (\ln x)' = \dfrac{1}{x\sqrt{4+x}} \]

Answer 6

Would it be 1/sqrt(4+e^ln(x)*1/x= 1/x*sqrt(4+x)?

Answer 7

Yep !

Answer 8

Thank you

Answer 9

Notice that \(f'(x)\) is never \(0\),
and x = 0 is a singularity - for x > 0, the function is strictly increasing

Answer 10

i think it deserves a better explanation, let me tag the experts @eliassaab @oldrin.bataku

Answer 11

For \(f\) to be invertible it must be injective; in other words, \(f\) must preserve distinctness so \(f(a)=f(b)\implies a=b\) (this should make sense -- convince yourself). For a continuous function, to ensure distinctness we require that our function is strictly monotone, either always increasing or always decreasing (you can visually see this on a graph using a "horizontal line test"). In other words, we must show that \(f'\) remains either non-negative or non-positive but never alternates in sign, so that it never "stands still".

Answer 12

stands still or repeats itself*

Answer 13

Now, we have: $$f(x)=\int_0^{\ln x}\frac1{\sqrt{4+e^t}}\,dt$$For convenience, let's consider the function \(g(u)=f(e^u)\) so as to simplify our integral bounds:$$g(u)=\int_0^u\frac1{\sqrt{4+e^t}}dt$$Note that no information is lost and the constrains on our domain are still \(x>0\). It is obvious that \(g\) is an antiderivative of \(1/\sqrt{4+e^u}\) so we can now differentiate both sides to find:$$\frac{dg}{du}=\frac1{\sqrt{4+e^u}}$$Now recall that \(g(u)=f(e^u)\) so \(dg/du=e^u (df/dx)(e^u)\), and coupled with the fact that \(e^u=x\) we have:$$x f'(x)=\frac1{\sqrt{4+x}}$$Clearly \(f'\) is only defined for \(x>0\) and it's readily apparent that \(\sqrt{4+x}>0\) for \(x>0\) so we've shown that \(f'\) never alternates in sign, and this is strictly monotone, so injective, and thus invertible.

Answer 14

my way is slightly more convoluted than @ganeshie8's but the general approach is the same. His point about using the fundamental theorem of calculus is this: define \(F\) as follows:$$G(t)=\int\frac1{\sqrt{4+e^t}}dt$$so that \(f\) can be written as follows$$f(x)=G(\ln x)-G(0)$$ergo \(f'(x)=G'(\ln x)\cdot\dfrac1x=\dfrac1{\sqrt{4+e^{\ln x}}}\cdot\frac1x=\dfrac1{x\sqrt{4+x}}\)

Answer 15

define \(G\) rather

Answer 16

You do not really need the derivative to show that f is increasing.
ln(x) is increasing so if
\[
a < b \implies \ln(a) < \ln(b)
\]
if h(t) is strictly positive for every t, then
\[
\int_0^{\ln(a)} h(t)dt < \int_0^{\ln(b)} h(t)dt
\]
You are integrating h(t) over a bigger interval

Answer 17

wow ! that observation makes it easy to visualize why area will be strictly increasing xD thank you both :)