I have a bit of elementary calculus problem for which my knowledge has become rusty.
I have the function $\left ( a-1 \right )x-ax^{3}$. I would like to determine the values of a for which the function $f\left ( x \right ) \in \left [ -1,1 \right ]$.
To do so, I begin with determining the extrema using the stationary point:
$f'\left ( x \right )=a-1-3ax^{2}=0$
This yields the two points $x_{\pm }=\pm \left [ \frac{1}{3}\left ( 1-\frac{1}{a} \right ) \right ]^{1/2}$ for which the extrema lies. $x_{+}$ corresponds the a maxima and $x_{-}$ corresponds to a minima of the function.
How do I continue?
My text mentioned that $0\leq 1-\frac{1}{a}<3$ for which I am unable to understand how it is arrived at and the principles behind.
Any explanation would be appreciated.
Well, at first note that for every $x\in\mathbb{R}$ it holds that: $$f(-x)=-f(x)\mbox{ and }f(0)=0$$ so, if there exists a $A>0$ such that for every $x\in[0,A]$ it holds that $|f(x)|\leq1$ for some values of $a$ then, for exactly the same values of $a$, it holds that $|f(x)|\leq1$ for $x\in[-A,0]$. Then, we can restrict our research to an interval of the form $[0,A]$, where $A>0$ and $A$ is independent of $a$.
Now, let us also make a crucial note: $$f(1)=a-1-a=-1\mbox{ and }f(-1)=-1+a-a=1$$ for every $a\in\mathbb{R}$, which means that a "convenient" choice of $A$ would be: $$A=1$$ Now, in the interval $[0,1]$, since $f$ is continuous, it attains both its maximum and its minimum. To find them, we have: $$f'(x)=a-1-3ax^2$$ and, then: $$f'(x)=0\overset{x\geq0}{\Leftrightarrow}x=\sqrt{\frac{a-1}{3a}}=x_0$$ which are well-defined iff $$\frac{a-1}{3a}\geq0\Leftrightarrow a< 0\mbox{ or }a\geq1$$ Now, we also need - since $x_0\in[0,1]$ - that $$0\leq x_0\leq1\Leftrightarrow\dots\Leftrightarrow0\leq\frac{a-1}{3a}\leq1\Leftrightarrow a\in\left[-\frac{1}{2},0\right)\cup[1,+\infty)$$ Now, calculating $f(x_0)$ we get: $$f(x_0)=\frac{2}{3\sqrt{3}}\sqrt{\frac{(a-1)^3}{a}}$$ Since we need $|f(x_0)|\leq1$, we have: $$\frac{(a-1)^3}{a}\leq\frac{27}{4}\Leftrightarrow\dots\Leftrightarrow a\leq4$$ (taking into consideration that $a\in\left[-\frac{1}{2},0\right)\cup[1,+\infty)$). So, finally: $$a\in\left[-\frac{1}{2},0\right)\cup[1,4]$$ Lastly, for every $a\in[0,1)$ we shall note that $$f'(x)\leq0$$ for every $x\in[0,1]$ so $f$ is decreasing, which means that: $$|f(x)|\leq\max\{|f(0)|,|f(1)|\}=1$$ So, we get that: $$a\in\left[-\frac{1}{2},4\right]$$ and $x\in[-1,1]$.