"Finding number of real solution of 4x−22+xcos⁡(ex)+1=0" was answered by our experts

River Dudley

Answered question

2022-04-08

Finding number of real solution of

4^{x} - 2^{2 + x} \cos (e^{x}) + 1 = 0

blogspainmarax6qi

Beginner2022-04-09Added 12 answers

Rewrite our equation in the following form:

f (x) = g (x)

, where

f (x) = \frac{1}{4} (2^{x} + 2^{- x})

and

g (x) = \cos {e^{x}}

We need

| f (x) | \leq 1

,which gives

\log_{2} (2 - \sqrt{3}) \leq x \leq \log_{2} (2 + \sqrt{3})

But for these values of x we see that g decreases and g is a concave function for

x \leq 0

By the way f is a convex function, decreases on

[\log_{2} (2 - \sqrt{3}), 0]

and increases on

[0, \log_{2} (2 + \sqrt{3})]

Thus, our equation has two roots maximum: one on

(\log_{2} (2 - \sqrt{3}), 0)

and one on

(0, \log_{2} (2 + \sqrt{3}))

.
Now, since f and g they are continuous functions,

f (\log_{2} (2 - \sqrt{3})) > g (\log_{2} (2 - \sqrt{3}))

f(0)

f (\log_{2} (2 + \sqrt{3})) > g (\log_{2} (2 + \sqrt{3}) t)

we see that our equation has two roots exactly.

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