WAIT, SORRY. I FAILED.
I meant WITHOUT T.P. form... As in, say... (x-3)(x+2)(x+4)
And I don't know calculus D:
Then why are you looking at inflection points? Points of inflection are pretty much meaningless without talking about derivatives.
Honestly, at this level all you'll need to know is the first rule of differentiation;
if y = a.x
n:. y' = n.a.x
n-1 and if there is no x at all, the term vanishes. (This also works with inverses, the power n just becomes negative is all)
Where you're diff. y w.r.t. x.
So, with the example you gave me, the easiest way to deal with it is expand out the brackets;
So,
(x-3)(x+4)(x+2)
goes to
(x
2-x-6)(x+4)
and then to
y = x
3 + 3x
2 -10x -24
Then just differentiate through.
So for the first term, x
3 goes to 3x
2, and by identical method for the rest, we get
y' = 3x
2 + 6x -10.
You'll notice the last term has dropped off; there was no x, so its derivative is 0.
Then you just do it again;
y" = 6x + 6
Then we set it equal to 0.*
Solving for x gives us x = -1.
HUZZAH!
Honestly man, don't stress too much about maths. In high school, I used to hate it, I'd average 65%, 70% if I really tried. When I got to uni though, it was explained differently, and the whole thing clicked. I jumped up to an average maths grade of 90%, and learned to love it.
So if you're struggling with it, don't worry. Keep at it, DO NOT STRESS, and eventually some little switch in your head will flick, and you'll be kicking yourself over how easy it all is.
*I should note at this point that while the inflection point always occurs at y" = 0, it's not always an inflection point when y" = 0. I.e. if it's an even powered polynomial (x
4 for example) the second derivative may touch y" = 0, but then curve away again without crossing. Basically, look for where the gradient in the original function y goes from curving up to curving down or vice versa.
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Topic: I fail at maths, a LOT (Read 3947 times)