Number of solution for an exponential equation

[juantheron]

The no. of real solution of $$4x^2 = e^x.$$

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

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

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

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

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

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

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

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

Thanks

 

[MarkFL]

Have you considered that you also have:

$$f(1)>0$$

$$f(5)<0$$

 

[juantheron]

Have you considered that you also have:

$$f(1)>0$$

$$f(5)<0$$

Thank you, MarkFL. Got it.

 

[chisigma]

The no. of real solution of $$4x^2 = e^x.$$

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

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

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

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

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

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

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

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.