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the function $f(x)=\sqrt{ax^3+bx^2+cx+d}$ has its non zero local minimum and local maximum values at -2 and 2 respectively.Given $'a'$ is the root of equation $x^2-x-6=0$ Doubt: solution given in book is: "since minimum occurs before maximum so $a<0$." Please explain this point.

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  • $\begingroup$ What domain is $f$ defined on, given that the expression under the radical can become negative? $\endgroup$ – dxiv Nov 16 '16 at 4:07
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Just consider $p(x) = ax^3 +bx^2+cx+d$

Now , $p'(x) = 3ax^2+2bx+c$, $p''(x) = 6ax+2b$

Now, roots of quadratic are $$p_{1,2} = \frac{-2b \pm \sqrt{4b^2-4(3a)(c)}}{6a} = \frac{-b \pm \sqrt{b^2-3ac}}{3a}$$

$$\implies p''(p_{1,2}) = \pm2\sqrt{b^2-3ac}$$

Hence, for minima to occur before maxima, we would need the algebraically smaller root to have a positive $p''(x)$, and vice versa. It is trivial to observe for that to happen, $a\lt0$

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$f$ has the same local maxima and minima with $g(x)=ax^3+bx^2+cx+d$, (possible less because of square root, and also have to pay attention at the introduced local maxima and minima at the boundary)

If $x_0$ is the local minimum, you will have that $g(x)>g(x_0)$ on a neighborhood of $x_0$ (so we can choose $x_1<x_0$ with $g(x_1)>g(x_0)$). If $a<0$ then $lim_{x\rightarrow\infty}g(x)=-\infty$, so you would find an $x_2<x_1<x_0$ such as $g(x_2)<g(x_1)>g(x_0)$ So there is a local maximum between $x_2$ and $x_0$ by continuity. But $g$ is a polynomial of degree 3, so it has at most a minimum and a maximum.

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