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Find the set of all the possible values of $a$ for which the function $f(x) = 5 + (a-2)x + (a-1)x^2 - x^3$ has a local minimum value at some $x < 1$ and local maximum value at some $x > 1$

The first derivative of f(x) is :

$f'(x) = (a-2) + 2x(a-1) -3x^2$

I do know the first derivative test for local maxima and local minima, but I can't figure out how I could use monotonicity to find intervals of increase and decrease of $f'(x)$

The expression for $f'(x)$ might suggest the double derivative test is the key, considering $f''(x) = 2(a-1) - 6x$ for which the intervals where it is greater than zero and less than zero can be easily found, but then again I can't think of a way how I could find a $c$ such that $f'(c) = 0$.

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  • $\begingroup$ I'm so sorry for the mistake. I've made the required edit. $\endgroup$ – skb Jul 13 '18 at 13:37
  • $\begingroup$ No problem, skb. I was just pointing it out because I figured it was probably a typo. (I make my share of typing mistakes, too!) $\endgroup$ – Namaste Jul 13 '18 at 13:43
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$f'(x) = 0$ is an inverse parabola, which has real roots in $x$ if

$$4[(a-1)^2 + 3(a-2)]\geq 0,$$ or, $$a \in \left[\frac{-1-\sqrt{21}}{2},\frac{-1+\sqrt{21}}{2}\right].$$

Let that be the case, then $f'(x)=0$ means that there are two critical points, namely

$$x_1 = \frac{a-1 + \sqrt{(a-1)^2+3(a-2)}}{3}, x_2 = \frac{a-1 - \sqrt{(a-1)^2+3(a-2)}}{3}.$$

Now, you need to check whether $f''(x_1)\lessgtr 0$ and $f''(x_2)\lessgtr 0$ as well as whether the critical points satisfy $x<1$ or $x>1$ as in your question.

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