# Finding the zeroes of a function

• Apr 19th 2009, 04:27 AM
gusztav
Finding the zeroes of a function
Hi,

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

I tried something along the lines of

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

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

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

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

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

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

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

Many thanks!
• Apr 19th 2009, 04:56 AM
stapel
I don't think the equation can be solved algebraically. I think you have to use numerical methods to find approximate zeroes.
• Apr 19th 2009, 05:07 AM
Moo
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)
$f'(x)=x+\frac 1x$
so $f'(x)>0$
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.