value of $a+b$ of the following function $f(x)=x^3-3x^2+5x\;$ and
$\;f(a)=1,f(b)=5.\;$ Find $a+b$.
I know only one real root exist for each equation as derivative of the function is always positive .I do not intend to use the formula of roots of cubic equation. How should i go about this problem ????
 A: I assume that $a$ and $b$ are real numbers (this can be guaranteed since $f(a)=1$ and $f(b)=5$ have real solutions.
As you said that $f^\prime$ is always positive that means $f$ is a strictly increasing function. Since $f(0)=0$ then $0<a<b$.
Note that we have
$$f(\frac{a+b}2)-\frac{f(a)}2-\frac{f(b)}2=-\frac38(b-a)^2(a+b-2).$$
So, If $a+b\leq 2$ then $$f(\frac{a+b}2)-\frac{f(a)}2-\frac{f(b)}2\ge0\iff f(\frac{a+b}2)\ge 3\iff\frac{a+b}2\ge 1\iff a+b\ge 2 $$
which gives $a+b=2$.
If $a+b>2$, then have $$f(\frac{a+b}2)-\frac{f(a)}2-\frac{f(b)}2<0\iff f(\frac{a+b}2)< 3\iff\frac{a+b}2< 1\iff a+b< 2 $$ that leads to contradiction.
In conclusion, we must have $a+b=2$.
A: Since the function $f(x)=x^{3}-3x^{2}+5x$ is increasing and $f(0)=0, f(1)=3$ and $f(2)=6$, we must have $0 \lt a \lt 1 \lt b \lt 2$. And also, we have, $$a^{3}-3a^{2}+5a=1 \space \space \space \space \space ...(1)$$ and $$b^{3}-3b^{2}+5b=5 \space \space \space \space \space ...(2)$$ Adding $(1)$ and $(2)$ and denoting $a+b$ and $ab$ by $t$ and $x$ respectively, we have, $$t^{3} -3t^{2}+5t -3xt+6x-6=0$$ $$\iff (t-2)(t^{2}-t+3-3x)=0.$$
Now, we have $t^{2} \geq 4x$ and hence we have $t^{2}-t+3-3x \geq x-t+3$. But $x-t+3 = (a-1)(b-1)+2 > -1 +2 =1 >0.$ This is because we have $-1 \lt -(b-1) \lt (a-1)(b-1)$. Hence we must have $$t=a+b=2.$$
