how to find the real root of

Question

zubhanwc3 · Answer

$$x^3-3x-6$$

phi · Answer

You could use a "root finding" algorithm such as newton-raphson

zubhanwc3 · Answer

i dont know that yet, im in calc bc, and if thats part of calc, we havent covered that yet

zubhanwc3 · Answer

the actual question i got was x^2-3 where 0<x<=b and 6/x where x>b
im supposed to find the value of b to make the function continuous
i set the function equal to each other, and got that function

zubhanwc3 · Answer

if u can explain the newton-raphson i could probably use that

phi · Answer

you could graph both functions and see where they intersect.

zubhanwc3 · Answer

i got the answer that way, but i want to be able to do it by hand :(

zubhanwc3 · Answer

so i can show my work by hand

phi · Answer

There is a closed form solution to cubics, but it is a bit complicated. https://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method according to wolfram, the real root is $$ x = 1/3 (81-54 \sqrt{2})^{1/3} +(3+2 \sqrt{2})^{1/3} $$

zubhanwc3 · Answer

thank you, after watching a video about that method, i must say, that method is so easy, just to confirm, it works for all cubics right?

phi · Answer

there is a closed form for solving all cubics.  If they are not in "depressed form" (i.e. missing the squared term) there is a procedure to make it depressed.  and then from there you use cardano or viete(sp?)

zubhanwc3 · Answer

just a question, at what lvl of math would u learn cardanos method? calc 2? or wat