Use quadratic formula to calculate x and then graph it to get max(f(x))=9,min(f(x))=2

Sprites wrote: Use quadratic formula to calculate x and then graph it to get max(f(x))=9,min(f(x))=2 This problem is not difficult but it's not so dull! Btw, it cannot be done like that, I highly doubt you did any calculations. You just put the expression of the bigger root into some online graph calculator and saw the result. The trick that can be used is very popular now, but maybe it was not so popular back then. $$(k^2+1)x^2+10kx-6(9k^2+1)=0 \qquad(1)$$Let $x_1\le x_2$ be the roots of $(1)$ (in case they are real). By Vieta we get $x_1x_2<0$ so one of the roots is negative and the other is positive. Therefore, the problem can be interpreted in the following way. Let $X$ be the set of all $x>0$ for which $(1)$ has a real solution with respect to $k.$ We are looking for $\min (X)$ and $\max (X)$ if they exist. We rewrite $(1)$ as $$(x^2-6\cdot 9)k^2 +10xk+x^2-6 =0\qquad (2)$$This equation has real solutions with respect to $k$ if and only if its discriminant is non negative $$10^2x^2-4(x^2-6)(x^2-6\cdot 9) \ge 0$$Setting $y=x^2$ we get $$-y^2+85y -6^2\cdot 9 \ge 0$$which yieds $y\in [4,81].$ Now, since $x>0$ we get $X=[2,9],$ thus $\min (X)=2, \max (X) =9.$