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While solving some questions related to functions, I came across this question and I am unable to find its range.

$$f(x)=3\cos^4x-6\cos^3x-6\cos^2x-3$$in the interval $[-π/2, π/2]$

I tried graphing this function on Desmos and got this result.

Desmos graph for f(x)

However, since I can't use a graphing tool in exams, I need to be able to solve this algebraically.

When I tried factorising the function, I could only get this far:

$$f(x)=3(\cos x+1)(\cos^3x-3\cos^2x+\cos x-1)$$

Can anyone help how to solve this further?

José Carlos Santos
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Navya
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2 Answers2

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Let $\varphi(x)=3x^4-6x^3-6x^2-3$; then $f(x)=\varphi\bigl(\cos(x)\bigr)$.

If $x\in[0,1]$, you have $\varphi'(x)\leqslant0$ and $\varphi'(x)=0\iff x=0$. So, $\varphi$ is decreasing on $[0,1]$. Since $\varphi(0)=-3$ and $\varphi(1)=-12$, the range of the restriction of $\varphi$ to $[0,1]$ is $[-12,-3]$ (see the graph of $\varphi$ below). Since the range of $\cos$ on $\left[-\frac\pi2,\frac\pi2\right]$ is $[0,1]$, the range of $f(=\varphi\circ\cos)$ is also $[-12,-3]$.

enter image description here

José Carlos Santos
  • 427,504
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Here is a solution which does not use differentiation.

Let $$c=\cos \theta \in [0,1] \ \text{because} \ \theta \in [-\pi/2,\pi/2]$$

The given expression is

$$\varphi(c)=3c^4-6c^3-6c^2-3$$

Let us first observe that $\varphi(0)=-3$ and $\varphi(1)=-12$.

Therefore, according to the intermediate value theorem, the range of $\varphi$ contains interval $[-12,-3]$.

Besides, let us express $\varphi(c)$ under the form:

$$\varphi(c)=3c^2((c-1)^2-3)-3 \tag{1}$$

As $0\le c \le 1$ , we have:

$$\begin{cases}-3&\le&(c-1)^2-3 &\le& -2\\ \ \ \ 0 &\le& \ \ \ \ \ \ \ 3c^2& \le & \ \ \ 3\end{cases}$$

implying:

$$-9 \le 3c^2((c-1)^2-3) \le 0$$

A final subtraction gives:

$$-12 \le \underbrace{3c^2((c-1)^2-3)-3}_{\varphi(c)} \le -3$$

showing that the range of $\varphi$ is now contained into interval $[-12,-3]$.

As a consequence, the range of $f$ is precisely $[-12,-3]$.

Jean Marie
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  • For what it's worth, I very much liked your (non-Calculus) approach, except for the last paragraph. Normally, such Topology concepts such as a compact interval or a continuous mapping are taught after Real Analysis, rather than before. Although it may be very difficult, is there any way that you can re-write your last paragraph so that the idea being expressed is crystal clear in the pre-Calculus world? – user2661923 May 28 '21 at 16:05
  • @ user2661923 I appreciate your well-founded remark. I am going to attempt it. You will say me if the objective is reached. – Jean Marie May 28 '21 at 17:11
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    +1 : nice editing. – user2661923 May 28 '21 at 17:49
  • @Navya Is my (re-edited) solution understandable at your level ? – Jean Marie May 28 '21 at 22:34
  • To Jean Marie : your last comment flagged me, instead of the original poster. For a comment that follows a specific answer or question, assuming that the addressee is not the originator of the answer or question, then the addressee is only flagged if they have registered for that question or answer, by first leaving a comment, following that question or answer. Since Navya is unregistered for your answer, in order to flag Navya, you should leave a comment that follows the original question. – user2661923 May 28 '21 at 22:45
  • I understand. Sorry. Thanks for this explanation. Besides, thanks again for having suggested me to re-write my answer in a simpler way. I appreciate very much this type of exchanges. – Jean Marie May 28 '21 at 22:50
  • @JeanMarie Yes! Thank you so much, I am able to understand it well. Thanks for linking the Intermediate value theorem wiki page :) – Navya May 29 '21 at 11:21