General 3rd degree polynomial always increasing problem?

[arpitm08]

General 3rd degree polynomial always increasing problem? Please Help!

What conditions on a, b, and c will make f(x)=ax^3+bx^2+cx+d always increasing?

For a function to be always increasing, the first derivative has to be always positive. So,

f1(x)=3ax^2+2bx+c>0

I tried finding the roots, that didn't lead me anywhere. Could someone please help?

I know that for a 3rd degree polynomial to be always increasing, it has to be a perfect cube. like (ex-f)^3, then you can expand that to be (ex)^3 -3f(ex)^2+3exf^2+f^3. Then since this will have to be the formula for f(x). So a=e^3, b=-3fe^2, c=3ef^2, d=f^3, but i don't know what to do from here. Please help!

 

[Mark44]

What conditions on a, b, and c will make f(x)=ax^3+bx^2+cx+d always increasing?

For a function to be always increasing, the first derivative has to be always positive. So,

f1(x)=3ax^2+2bx+c>0

I tried finding the roots, that didn't lead me anywhere. Could someone please help?

I know that for a 3rd degree polynomial to be always increasing, it has to be a perfect cube. like (ex-f)^3, then you can expand that to be (ex)^3 -3f(ex)^2+3exf^2+f^3. Then since this will have to be the formula for f(x). So a=e^3, b=-3fe^2, c=3ef^2, d=f^3, but i don't know what to do from here. Please help!

I think I finally figured out what you meant by (ex-f)^3. It would have been easier to grasp if you had written (Ax - B)^3, since e and f already have other meanings -- the natural number e, and f as in f(x).

Some things to think about.
a should be positive.
There should be exactly 1 real root. (If there were 3 roots, the graph would have something of an S shape. If there were 2 roots, one root would be repeated, and the graph would drop down and touch the x-axis rather than cross it.)
See if you can make up equations for 3rd degree polynomials with 1 root, 2 roots, 3 roots.
f'(x) should be >= 0 for all x.

What about f''(x)? Can f''(x) change sign? If so, the concavity is changing. If so, how many times can the concavity change sign?

 

[HallsofIvy]

What conditions on a, b, and c will make f(x)=ax^3+bx^2+cx+d always increasing?

For a function to be always increasing, the first derivative has to be always positive. So,

f1(x)=3ax^2+2bx+c>0

I tried finding the roots, that didn't lead me anywhere. Could someone please help?

I wish you had shown what you did. The derivative is a quadratic so will always be positive if and only if a is positive and the quadratic is never 0. That will happen when the discrimant is negative: that is if (2b)²- 4(3a)(c)= 4b2- 12ac< 0 which is the same as b²< 3ac.

The cubic f(x)= ax³+ bx²+ cx+ d is always increasing if and only if a> 0 and b²< 3ac. It is always decreasing if and only if a< 0 and b²< 3ac.

I know that for a 3rd degree polynomial to be always increasing, it has to be a perfect cube. like (ex-f)^3, then you can expand that to be (ex)^3 -3f(ex)^2+3exf^2+f^3. Then since this will have to be the formula for f(x). So a=e^3, b=-3fe^2, c=3ef^2, d=f^3, but i don't know what to do from here. Please help!