# Increasing and decreasing intervals

Find a polynomial $f(x)$ of degree 4 which increases in the intervals $(-∞,1)$ and $(2,3)$ and decreases in the intervals $(1,2)$ and $(3,∞)$ and satisfies the condition $f(0)=1$

It is evident that the function should be $f(x)=ax^4+bx^3+cx^2+dx+1$. I differentiated it. Now, I'm lost. I tried putting $f'(1)>0$, $f'(3)-f'(2)\geq 0$, and $f'(2)-f'(1)\leq 0$. Am I doing correct? Or is there another method?

• set 1, 2, and 3 as local maximum or minimum. – user295959 Aug 5 '18 at 7:08
• @user295959 thanks for the help! – Arka Seth Aug 5 '18 at 7:20

According to the given conditions, we may assume $$f(x)=ax^4+bx^3+cx^2+dx+1.$$

Then $$f'(x)=4ax^3+3bx^2+2cx+d.$$ Notice that $x=1,2,3$ are the roots for $f'(x)=0$. Hence

$$f'(x)=4ax^3+3bx^2+2cx+d=k(x-1)(x-2)(x-3)=kx^3-6kx^2+11kx-6k.$$

Thus, $$a=\frac{k}{4},~~~b=-2k,~~~c=\frac{11}{2}k,~~~d=-6k.$$

Therefore, we obtain $$f(x)=\frac{k}{4}x^4-2kx^3+\frac{11k}{2}x^2-6kx+1,$$ where $k>0.$

Since $1,2,3$ are zeroes for $f'$ we have $$f'(x) = a(x-1)(x-2)(x-3) = ax^3-6ax^2+11ax-6a$$

So, by integrating we get $$f(x) = {a\over 4}x^4 -2ax^3-{11a\over 2}x^2-6ax +c$$

Puting $f(0)=1$ we get $c=1$ and thus:$$f(x) = {a\over 4}x^4 -2ax^3-{11a\over 2}x^2-6ax +1$$