Simple equation solving (I think)

  • Thread starter Thread starter tachu101
  • Start date Start date
AI Thread Summary
To find the x-intercepts of the function f(x) = ln(x+1) - (sinx)^2, the equation can be simplified to ln(x+1) = (sinx)^2. It is noted that there are three x-intercepts, with x = 0 being the most apparent. Analyzing the maximum and minimum values of sin^2(x) helps establish a range of x-values where x-intercepts may exist. Graphing f(x) within this domain allows for estimation of the x-intercepts. While Newton's method can refine these estimates, it may be considered complex for a pre-calculus level.
tachu101
Messages
74
Reaction score
0

Homework Statement



Find x-intercepts of f(x)= ln(x+1)-(sinx)^2

I know to get to ln(x+1)=(sinx)^2 but have no idea what to do after that.
 
Physics news on Phys.org
I don't think it's THAT easy, x = 0 comes to mind though
 
It turns out that there are 3 x-intercepts (x=0 is the obvious one). You can estimate what domain they lie in by looking at the maximum and minimum values of \sin^2(x)...what are those? What does that tell you about the max/min of \ln(1+x) for which there might be any x-intercepts? You can use that to determine a range of x-values for which x-intercepts are possible.

The next step would be to graph f(x) over that Domain and estimate value for the x-intercepts.

If you are familiar with Newton's method, you can improve your estimations through a few iterative calculations.
 
This seems to be kind of an overkill for someone in pre-calc though unless you were allowed to use a CAS.
 
Use iteration?
 
Yes you could use Newton's as was mentioned but I still maintain it's a somewhat hard problem for pre-calc then again I never took pre-calc and maintain that it's a useless class :)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Back
Top