the real number $s$ lies in the interval
(A)$(-0.75,-0.5)$
(B)$(-0.5,0)$
(C)$(0,1)$
(D)$(-0.25,0)$
and the area of region bounded by $f(x),y=0,x=0$ lies in the interval
(A)$(0.75,3)$
(B)$(0,\frac{21}{64})$
(C)$(\frac{21}{64},\frac{11}{16})$
(D)$(9,10)$
attempt
$f'(x)$ has no roots. $x=-0.25$ is a point of inflection. The root lies in $(-1,-0.5)$.
Hence (A) is the correct answer.
For second part, the area is $t+t^2+t^3+t^4$.
Its minimum value is when $f(t)=0$. And since A is correct for first part, $t\in(0.75,0.5)$, the Maximum value is when $t=0.75$ and minimum value is when $t$ is the root of the actual equation. How do I find that?
You are correct that $f'(x)$ has no (real) roots, but you've miscalculated the point of inflection. Also, there's no particular purpose (that I can think of) to calculating the point of inflection, in the first place. Instead, one could easily use the Intermediate Value Theorem on $(-0.75,-0.5),$ since $f$ has only one real root (why?).
Also, $(0.75,0.5)$ makes no sense, and neither does "$f(t)=0.$" Rather, we know that $t\in(0.5,0.75),$ so letting $g(x)=1+x+x^2+x^3+x^4,$ we have that $$g(0.5)<g(t)<g(0.75).$$ (Can you see why?) Can you take it from here?