Let $f(x)=2\arccos x+4\operatorname{ arccot } x-3x^2-2x+10, x\in[-1,1]$. If $[a,b]$ is the range of $f(x)$, find $4a-b$.

Question

Question:

Let $f(x)=2\arccos x+4\operatorname{ arccot } x-3x^2-2x+10, x\in[-1,1]$. If $[a,b]$ is the range of $f(x)$, find $4a-b$.

Method $1:$

$f'(x)=-\frac{2}{\sqrt{1-x^2}}-\frac{4}{1+x^2}-6x-2$

1st, 2nd and 4th terms are always negative. But not 3rd. Thus, can we say $f'(x)$ is negative?

Method $2:$

$\arccos x, \operatorname{ arccot } x, -2x$ are always decreasing. But not $-3x^2$. Thus, can we say $f(x)$ is decreasing?

Answer 1

The first thing that is required to be found in the domain of the function which is $[-1,1]$ as we have to satisfy the domain of $\arccos(x)$ That done, we proceed to find the derivative, $$f′()=−\frac2{\sqrt{1−^2}}-\frac4{1+x^2} -6x-2$$ when $x$ is between $0$ and $1$, the derivative is negative indicating a decreasing function in that portion. when $x$ is between $-1$ and $0$, we see that though the $-6x$ term is positive, it is still smaller in magnitude than the rest of the terms. (this you can confirm with any arbitrary value) so the function is a decreasing function in its domain and it is also injective. now all we have to do is find the values of $f(x)$ at $x=-1$ and $x=+1$ to get the range.