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Let $f(x)=ax^3+bx^2+cx+d$, where $a,b,c,d$ are real is an increasing function. Given that $3b^2<c^2$. Define $g(x)= af’(x) + bf’’(x) + c^2$. Also , define $p(x) = \int_x^m g(t)dt$. Classify $p(x)$ as an increasing , or decreasing , or neither increasing nor decreasing function. (Given that $m$ is a real number)

This is how I attempted it :

We’re given that $f(x)$ is an increasing function . Therefore it’s derivative must always be greater than zero. Hence ,

$3ax^2 + 2bx +c >0$

Now , this forms a quadratic equation. And if it is always greater than zero, then it’s discriminant must be always less than zero.

Thus we will have ,

$4b^2-12ac<0$ or

$b^2<3ac. (1)$

Also we have been given that $3b^2<c^2 (2)$

Now , by integration , we obtain $p(x)$ as

$p(x) = a^2x^3 - a^2m^3 + 4abx^2 - 4abm^2 + acx - acm + 2b^2x - 2b^2m + c^2x- c^2m$

Now the derivative of $p(x)$ would be

$\frac{d}{dx} p(x) = 3a^2x^2 -8abx +ac +2b^2 + c^2$

This again forms a quadratic in $x$. If the discriminant of the above quadratic is less than zero, it would always be positive because the coefficient of $x^2$ will always be positive.

Therefore , we find that the discriminant of the above quadratic is

$64a^2b^2 - 4(3a^2)[ac+2b^2+c^2]$

$=40a^2b^2-12a^3c-12a^2c^2$

We first assume that it really is always less than zero , therefore we get an inequality i.e

$40a^2b^2<12a^3c +12a^2c^2$

Now if we are able to prove this inequality we will get that the discriminant is always negative which means that the derivative is always positive . So the function is strictly increasing.

From equation $1$

$b^2<3ac$ , therefore $4b^2<12ac$

And from the second equation we get

$3b^2<c^2$ thus $36b^2<12c^2$

Adding the above inequalities , we would get $40b^2<12ac + 12c^2$ which proves the desired result . So $p(x)$ is strictly increasing.

Do you think my reasoning is correct? Because according to my book the function is neither increasing nor decreasing. Where do you think have I made a mistake? Please correct me if I have . Thanks for your help !

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  • $\begingroup$ If you look up the general formula for cubic equations, you should find all you need to determine what is increasing or decreasing. $\endgroup$ – poetasis May 21 '18 at 17:10
  • $\begingroup$ Try math.vanderbilt.edu/schectex/courses/cubic . I used it and it gave me the answers I needed. $\endgroup$ – poetasis May 21 '18 at 17:16
  • $\begingroup$ @poetasis : thanks ! But do you find any mistakes in this ? $\endgroup$ – Aditi May 21 '18 at 18:12
  • $\begingroup$ The discriminant will tell you if it's non-increasing or non-decreasing depending on whether it's positive, negative or zero. I don't have time to spend but it's in the links. Good luck. $\endgroup$ – poetasis May 22 '18 at 16:51

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