If the slope of the tangent to the curve is maximum at x = 1 and curve has minim value 1  at x = 0, then the curve which also satisfies the equation  $\frac{d^{3}y}{d{x}^3}=4x-3$ is
 

we have  $\frac{d^{3}y}{d{x}^3}=4x-3$
 Integrating, we get   $\frac{d^{2}y}{d{x}^2}=2x^2-3x+c$
 Given slope of tangent to the curve is maximum at x = 1,

$slope is \frac{dy}{dx}, to find slope max equate \frac{d^{2}y}{d{x}^2} to zero$
 Then  ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=1}=0\Rightarrow 2-3+c=0\Rightarrow c=1$
$\frac{d^{2}y}{d{x}^2}=2x^2-3x+1$  
 Integrating we get 
$\frac{dy}{dx}=\frac{2x^3}{3}-\frac{3x^2}{2}+x+D$  
$\therefore$curve has a minimum value 1 at x = 0

$to find y=f\left(x\right) minimum equate f^{\text{'}}\left(x\right)=0 or \frac{dy}{dx}=0$
$\therefore {\left(\frac{dy}{dx}\right)}_{x=0}=0\Rightarrow D=0$  
$\therefore \frac{dy}{dx}=\frac{2x^3}{3}-\frac{3x^2}{2}+x$  
 Again integrating, we get   $y=\frac{x^4}{6}-\frac{x^3}{2}+\frac{x^2}{2}+E$
 Given minimum value of curve is 1 at x = 0
 Then 1 = 0 + E  $\Rightarrow$  E = 1
 Hence the curve is   $y=1+\frac{x^2}{2}-\frac{x^3}{2}+\frac{x^4}{6}$