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A curve with equation , y=ax4+bx3+cx2+d , has zero gradient at point (0,1) and also touches the x-axis at point (-1,0).Then value of x for which the curve has a negative gradient are given by;ans is for x<-1plzz explain

Ranjita yadav , 12 Years ago

Grade 11

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
$y = ax^{4} + bx^{3} + cx^{2} + d$
$y' = 4ax^{3} + 3bx^{2} + 2cx$
$y' = x(4ax^{2} + 3bx + 2c)$
There are three roots of y’ = 0
Two roots are given: x =0, -1
$y = ax^{4} + bx^{3} + cx^{2} + d$
$1 = a.0^{4} + b.0^{3} + c.0^{2} + d$
$d = 1$
$0 = a.-1^{4} + b.-1^{3} + c.-1^{2} + d$
$0 = a - b + c + 1$
$b-c-a = 1$
Let x₁be the 3^rdroot of y’ = 0
$y' = x(4ax^{2} + 3bx + 2c) < 0$
$x(4ax^{2} + 3bx + 2c) < 0$
$a'x(x+1)(x-x_{1}) < 0$
a’ > 0
By wavy curve method, we have
x₁> 0
$\Rightarrow x < -1, x > x_{1}$

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