A curve with equation of the form $y=a x^4+b x^3+c+c x+d$ has zero gradient at the point (0, 1) and also touches the x-axis at the point (-1, 0) then the values of the x for which the curve has a negative gradient are:

$x<-1$

We have, $\frac{d y}{d x}=0$ at (0, 1) and (-1, 0)

∴  c = 0 and $-4 a+3 b-c=0 \Rightarrow a=\frac{3 b}{4}$ and c = 0             ......(i)

Also, the curve passes through $(0,1)$ and $(-1,0)$

∴  $d=1$ and $0=a-b-c+d \Rightarrow a-b-c+1=0$               ........(ii)

From (i) and (ii), we get

$a=3, b=4, c=0 \text { and } d=1$

∴  $y=3 x^4+4 x^3+1 \Rightarrow \frac{d y}{d x}=12 x^3+12 x^2$

Now,

$\frac{d y}{d x}<0 \Rightarrow 12 x^3+12 x^2<0 \Rightarrow x+1<0 \Rightarrow x<-1$