A worthy try, but then you'd be dividing the k part by x as well, and so the result isn't really a quadratic (because then you have an x to the -1 power, the one under the k).
With the leading coefficient positive, there's a third case where the derivative's always positive, so there's only one root. I can see how to tell whether this is true (looking at the discriminant of the derivative?), but is it reasonably easy to solve for h then?
In the extremes, the derivative sign will always be the same as the sign on the leading coefficient, for odd-degree polynomials. The derivative will switch signs at the stationary points. For a cubic, there are only two stationary points at most. This means, if there are stationary points, the derivative at the inflection point will be opposite the sign of the derivative at the extreme values of x. So this means, you just need to find the sign of the derivative at the inflection point, to determine if it is a "roller coaster cubic" with the two stationary points, or if it is a stretched out cubic with the same sign of its slope everywhere. If the derivative at the inflection point is zero, this means it is just the simple cubic (y=x^3), that is scaled and shifted. To find the inflection point, we need to find where the 2nd derivative is zero. For a general cubic y=a*x^3 + b*x^2 + c*x + d, this is determined by x = -b/(3*a).
You definitely have a gift of teaching. Keep it up 🙌
Wow ❤
I wish you were my maths teacher.
Very nice Sir.
Sir is there any easier method for getting the roots of a general cubic polynomial
great video ,god bless you
thanks a lot!
If u divide it by a a factor then get a quadratic can you use the discriminant
A worthy try, but then you'd be dividing the k part by x as well, and so the result isn't really a quadratic (because then you have an x to the -1 power, the one under the k).
With the leading coefficient positive, there's a third case where the derivative's always positive, so there's only one root. I can see how to tell whether this is true (looking at the discriminant of the derivative?), but is it reasonably easy to solve for h then?
In the extremes, the derivative sign will always be the same as the sign on the leading coefficient, for odd-degree polynomials. The derivative will switch signs at the stationary points. For a cubic, there are only two stationary points at most. This means, if there are stationary points, the derivative at the inflection point will be opposite the sign of the derivative at the extreme values of x.
So this means, you just need to find the sign of the derivative at the inflection point, to determine if it is a "roller coaster cubic" with the two stationary points, or if it is a stretched out cubic with the same sign of its slope everywhere. If the derivative at the inflection point is zero, this means it is just the simple cubic (y=x^3), that is scaled and shifted.
To find the inflection point, we need to find where the 2nd derivative is zero. For a general cubic y=a*x^3 + b*x^2 + c*x + d, this is determined by x = -b/(3*a).
Thanks a lot sir u helped me today
I don't know in which grade ur teaching this...but in India we face these questions in 12 grade
in england year 12. (Age 16-17). WHat age group in India?
Grade 12 in South Africa too. (Age 17-18)
12 grade in greece too
In India syllabus updated we do it in 8th grade....
Age group 11-14
number 1