A parabola passing through critical points of a curve!

Calculus Level 5

Let f ( x ) = x 6 6 x 2 + 6 x 7 f(x)=x^6-6x^2+6x-7 . It is known that this polynomial has three critical points. The equation of the parabola that passes through these three critical points is y a = b x 2 + c x + d . y^a = bx^2+cx+d. Find a + b + c + d . \mid a+b+c+d \mid .

Note : Critical points are defined as points where tangent is parallel to x x axis.


The answer is 5.

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2 solutions

Pranjal Jain
Nov 17, 2014

Let f ( x ) = x 6 6 x 2 + 6 x 7 f(x)=x^{6}-6x^{2}+6x-7 f ( x ) = 6 x 5 12 x + 6 \Rightarrow f'(x)=6x^{5}-12x+6

For critical point, f ( x ) = 0 x = 1 , y = 6 f'(x)=0\Rightarrow x=1, y=-6

Substituting x=1, y=-6 in y a = b x 2 + c x + d y^{a}=bx^{2}+cx+d , we will get b + c + d = 6 b+c+d=-6 .

For the equation to be a parabola, a=1.

Hence, a + b + c + d = 5 a + b + c + d = 5 a+b+c+d=-5\Rightarrow |a+b+c+d|=\boxed{5}

Question looks dirty initially but on getting x=1 as critical point, it does not look like a level 5 question! Still nice question!

Pranjal Jain - 6 years, 6 months ago

Its possible to find whole polynomial, without doing any big calculations. I could have asked the problem in better way :( . Try to find whole equation., then only the purpose of the problem will be served .

Shivang Jindal - 6 years, 6 months ago
Yashas Ravi
Feb 26, 2021

If f ( x ) = x 6 6 x 2 + 6 x 7 f(x) = x^6-6x^2+6x-7 , this means f ( x ) = 6 x 5 12 x + 6 f'(x)=6x^5-12x+6 . To find the critical points, f ( x ) = 0 f'(x)=0 , so x 5 2 x + 1 = 0 x^5-2x+1=0 meaning that x 5 = 2 x 1 x^5=2x-1 . Since the equation represents a parabola, a = 1 a=1 , so y = b x 2 + c x + y=bx^2+cx+ d. Since f ( x ) f(x) also contains these points, we know that x 6 6 x 2 + 6 x 7 = b x 2 + c x + d x^6-6x^2+6x-7=bx^2+cx+d . Since x 5 = 2 x x^5=2x- 1, this means x 6 = 2 x 2 x x^6=2x^2-x , so 2 x 2 x + 6 x 7 = b x 2 + c x + d 2x^2-x+6x-7=bx^2+cx+d . Simplifying yields y = 4 x 2 + 5 x 7 y=-4x^2+5x-7 , so the answer is 1 4 + 5 7 = 5 |1-4+5-7| = 5 which is the final answer.

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