Number of solution for an exponential equation

[juantheron]

The no. of real solution of \(\displaystyle 4x^2 = e^x.\)

\(\displaystyle \bf{My\; Try::}\) Let \(\displaystyle f(x) = 4x^2-e^x\;,\) Then \(\displaystyle f'(x)=8x-e^x\)

and \(\displaystyle f''(x)=8-e^x\) and \(\displaystyle f'''(x)=-e^x<0\; \forall x \in \mathbb{R}\)

So \(\displaystyle f'''(x) = 0\) has no real roots, So Using \(\displaystyle \bf{IVT}\)

equation \(\displaystyle f''(x) = 0\) has at most one real roots.

and equation \(\displaystyle f'(x) = 0\) has at most two real roots.

and equation \(\displaystyle f(x) = 0\) has at most three real roots.

But I am getting only one root which lie in \(\displaystyle (-1,0)\) because \(\displaystyle f(-1)>0\) and \(\displaystyle f(0)<0\)

So please help me How can I calculate other two roots.

Thanks

 

[MarkFL]

Have you considered that you also have:

\(\displaystyle f(1)>0\)

\(\displaystyle f(5)<0\)

 

[juantheron]

Have you considered that you also have:

\(\displaystyle f(1)>0\)

\(\displaystyle f(5)<0\)

Thank you, MarkFL. Got it.

 

[chisigma]

The no. of real solution of \(\displaystyle 4x^2 = e^x.\)

\(\displaystyle \bf{My\; Try::}\) Let \(\displaystyle f(x) = 4x^2-e^x\;,\) Then \(\displaystyle f'(x)=8x-e^x\)

and \(\displaystyle f''(x)=8-e^x\) and \(\displaystyle f'''(x)=-e^x<0\; \forall x \in \mathbb{R}\)

So \(\displaystyle f'''(x) = 0\) has no real roots, So Using \(\displaystyle \bf{IVT}\)

equation \(\displaystyle f''(x) = 0\) has at most one real roots.

and equation \(\displaystyle f'(x) = 0\) has at most two real roots.

and equation \(\displaystyle f(x) = 0\) has at most three real roots.

But I am getting only one root which lie in \(\displaystyle (-1,0)\) because \(\displaystyle f(-1)>0\) and \(\displaystyle f(0)<0\)

So please help me How can I calculate other two roots.

Thanks

It's interesting to analyse the more general case...

$\displaystyle a\ x^{2} = e^{x}\ (1)$

Extracting the square root we obtain from (1)...

$\displaystyle x\ e^{- \frac{x}{2}} = \pm \frac{1}{\sqrt{a}}\ (2)$

... the solution of which is...

$\displaystyle x = - 2\ W(\pm \frac{1}{2\ \sqrt{a}})\ (3)$

... where W(*) is the Lambert Function, which is represented in the figure...

W(*) is a multivalued function and for $- \frac{1}{e} < x < 0$ for each value of x we have two values of W(x). the conclusion is that (1) for $a> \frac{e^{2}}{4}$ has three real solutions, for $a= \frac{e^{2}}{4}$ two real solutions and for $a< \frac{e^{2}}{4}$ one real solution...

Kind regards

$\chi$ $\sigma$

 

[juantheron]

Thanks chisigma for Yours Nice generalization.