How would I go about finding the roots of (ex−1)−karctan⁡(x)=0?

Irrerbthist6n

Irrerbthist6n

Answered

2022-01-14

How would I go about finding the roots of (ex1)karctan(x)=0?

Answer & Explanation

reinosodairyshm

reinosodairyshm

Expert

2022-01-16Added 36 answers

If k>0
Let
f(x)=ex1ktan1xf(x)=exk1+x2,f(x)=ex+k(1+x2)2>0
f′(x) is an increasing function f(x)>f()>0f(x) is an increasing function. Hence f(x)=0, will have at most one real root. As f()=1+kπ2 and f()>0, for one real root by IVT f()<0. Finally this gives 0<k<2π. The one root is x=0.
For k>2π, both f()> and f()>0 so f(x)=0 will have 0,2,4,... number of real roots. Since x=0 is essentially a root. So there will two real roots if k>2π. Further, since f′′(x)>0, for the function f(x) can have atmost one min, this rules out more than two real roots.
RizerMix

RizerMix

Expert

2022-01-20Added 437 answers

Of course x=0 is always a solution. Otherwise, write the equation as k=ex1arctan(x) Call the right side f(x). The singularity at x=0 is removable, with limx0f(x)=1. We also have limxf(x)=2π and limx+f(x)=+ It appears that f(x) is increasing. If so, for 2π<k<1 and 1<k< there are two real roots (x=0 and the root of f(x)=k), otherwise there is only x=0.
alenahelenash

alenahelenash

Expert

2022-01-24Added 366 answers

Let f(x)=ex1karctan(x). We have f(0)=0. If k=0, obviously x=0 is the only solution. If k0, then f(x)=exk1+x2, so that f(x)=0(1+x2)ex=k. Since (1+x2)ex is increasing, goes to 0 at and goes to  at  we see that f′ has exactly one real zero, say f′(x0)=0. We see that f decreases on (,x0) and increases on (x0,), so that f attains its minimum at x0 Since we know that f(0)=0, it must be that x0<0 and x=0 is the only solution.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?