Question

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

• A. x > -1
• B. x < 1
• C. x < -1
• D. $-1\le \times \le 1$

Answer 1

The correct option is C x < -1

Given, $y=a{x}^4+b{x}^3+cx+d$
$\Rightarrow y^{\prime }=4a{x}^3+3b{x}^2+c$
Using given conditions,
$y\left(0\right)=1\Rightarrow d=1$
$y^{\prime }\left(0\right)=0\Rightarrow c=0$
$y(-1)=0\Rightarrow a-b=-1..\left(1\right)$
and $y^{\prime }(-1)=0\Rightarrow 4a-3b=0..\left(2\right)$
Solving equation (1) and (2) we get, $a=3,b=4$
Hence the polynomial is,
$y=3x^4+4x^3+1$
$y^{\prime }=12x^2(1+x)$
Now for negative gradient
$y^{\prime }<0\Rightarrow 12x^2(1+x)<0$

$\Rightarrow x<-1$