# Thread: Finding the zeroes of a function

1. ## Finding the zeroes of a function

Hi,

I'd really appreciate some pointers on how to find the zeroes of the function $\displaystyle f(x)=\frac{x^2}{2}+\log(x)$,
that is, $\displaystyle x \in <0,+\infty>$ for which $\displaystyle \frac{x^2}{2}+\log(x)=0$.

I tried something along the lines of

$\displaystyle \frac{x^2}{2}+\log(x)=0$

$\displaystyle x^2+2\log(x)=0$

$\displaystyle x^2+\log(x^2)=0$

$\displaystyle \log(e^{x^2})+\log(x^2)=0$

$\displaystyle \log(e^{x^2} \cdot x^2)=0$

$\displaystyle e^{x^2} \cdot x^2=1$

but don't really know how to proceed from that.

Many thanks!

2. I don't think the equation can be solved algebraically. I think you have to use numerical methods to find approximate zeroes.

3. Hello,

It is not possible to find the exact value for which it is 0.
Generally, equations involving both polynomials and logarithms are hard to handle.

You can use approximation methods (euler, bisection...) or find the derivative and study the function to get a general idea.

f is defined over (0,infinity)
$\displaystyle f'(x)=x+\frac 1x$
so $\displaystyle f'(x)>0$
this function is strictly increasing.
Limit to 0 is -infinity, f(1)=1/2
so there is a unique value a for which f(a)=0 and a is in (0,1)

but if you can find c and d such that f(c)<0 and f(d)>0, then for sure, a is between c and d.