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 ?

The mean value theorem does not tell you that every $c$ has a corresponding $a$ and $b$ with $\frac{f(a)-f(b)}{a-b}=f'(c)$.  It tells you that given $a$ and $b$ you can find a $c$, but it doesn't say given $c$ you can find $a$ and $b$.  In your case,
$\dfrac{f(a)-f(b)}{a-b}=4(a^2+ab+b^2)+3(a+b)-1=\dfrac{4(2a+2b+1)^2+(4a+1)^2+(4b+1)^2}{8}-\dfrac{7}{4}$.
The minimum of this thus occurs only for $a=b=-\frac{1}{4}$, but this is not allowed because then $a-b$ is $0$ in the denominator.  So there is no such $a$ and $b$ that gives you $-\frac{7}{4}$.
