For $f(x)=4x^3+3x^2-x-1$, The Range of values $\frac{f(x_1)-f(x_2)}{x_1-x_2}$ can take is- My Attempt :-
$f'(x)=12x^2+6x-1$ where $f'(x) \ge \frac{-7}{4}$. So (I think) from LMVT we can directly say that 
$$\frac{f(b)-f(a)}{b-a} \ge \frac{-7}{4}$$  But the answer Given is $$\frac{f(b)-f(a)}{b-a} > \frac{-7}{4}$$
So my Question is Is This Application of LMVT correct ,if yes then why $\frac{-7}{4}$ is excluded from the given range ?

Intuitively, as long as $f'(x)$ attains its minimum at one single point, there won't exist two distinct points where the secant matches that slope.
To formalize, assume $f'(x)$ has a global minimum $m$, then it follows from MVT that:
$$\frac{f(b)-f(a)}{b-a} \ge m \quad \text{for} \;\;\forall a \ne b$$
If $a,b$ points exist where the equality holds, then by MVT for the intervals $(a,\frac{a+b}{2})$ and $(\frac{a+b}{2},b)$:
$$m = \frac{f(b)-f(a)}{b-a} = \frac{1}{2}\left(\frac{f(b)-f(\frac{a+b}{2})}{\frac{b-a}{2}} + 
\frac{f(\frac{a+b}{2})-f(a)}{\frac{b-a}{2}}\right) \ge \frac{1}{2}(m+m) = m
$$
It follows that the middle inequality must be an equality as well, so:
$$m = \frac{f(b)-f(a)}{b-a} = \frac{f(b)-f(\frac{a+b}{2})}{\frac{b-a}{2}} =  \frac{f(\frac{a+b}{2}) - f(a)}{\frac{b-a}{2}}
$$
From which:
$$f\left(\frac{a+b}{2}\right) = \frac{f(a) + f(b)}{2}$$
It follows that $f(x)$ is both midpoint convex and concave, thus linear by continuity (see for example Function that is both midpoint convex and concave). But the given $f(x)$ is a $3^{rd}$ degree polynomial which is obviously not linear, so such points $a,b$ do not exist. Therefore the strict inequality holds.
