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$f(x)=x^3-3x^2+5x\;$ 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 ????

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  • $\begingroup$ Solving approximately with a graph, we can see that $a+b=2$. That may help reverse engineer it... $\endgroup$ – Zook May 7 '14 at 18:11
  • $\begingroup$ $a,b \in \Bbb R$? $\endgroup$ – Indrayudh Roy May 7 '14 at 18:13
  • $\begingroup$ Yup it is real , nice solution indrayudh $\endgroup$ – avz2611 May 7 '14 at 20:06
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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 $a+b=2$.

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  • $\begingroup$ How did you know to solve it like this? Is there a broader technique you are pulling from? $\endgroup$ – Zook May 7 '14 at 19:32
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    $\begingroup$ Fist I considered that $x=1$ is the point in which $f$ change for concav to convex. And since $a<1<b$ then $f(\frac{a+b}2)$ must lie around $\frac{f(a)+f(b)}2$. Then I proved analytically like above. $\endgroup$ – Jlamprong May 7 '14 at 19:37
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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 $$t=a+b=2.$$

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