A parabola passing through critical points of a curve!
Let f ( x ) = x 6 − 6 x 2 + 6 x − 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 . Find ∣ a + b + c + d ∣ .
Note : Critical points are defined as points where tangent is parallel to x axis.
The answer is 5.
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2 solutions
Question looks dirty initially but on getting x=1 as critical point, it does not look like a level 5 question! Still nice question!
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 .
If f ( x ) = x 6 − 6 x 2 + 6 x − 7 , this means f ′ ( x ) = 6 x 5 − 1 2 x + 6 . To find the critical points, f ′ ( x ) = 0 , so x 5 − 2 x + 1 = 0 meaning that x 5 = 2 x − 1 . Since the equation represents a parabola, a = 1 , so y = b x 2 + c x + d. Since f ( x ) also contains these points, we know that x 6 − 6 x 2 + 6 x − 7 = b x 2 + c x + d . Since x 5 = 2 x − 1, this means x 6 = 2 x 2 − x , so 2 x 2 − x + 6 x − 7 = b x 2 + c x + d . Simplifying yields y = − 4 x 2 + 5 x − 7 , so the answer is ∣ 1 − 4 + 5 − 7 ∣ = 5 which is the final answer.
Let f ( x ) = x 6 − 6 x 2 + 6 x − 7 ⇒ f ′ ( x ) = 6 x 5 − 1 2 x + 6
For critical point, f ′ ( x ) = 0 ⇒ x = 1 , y = − 6
Substituting x=1, y=-6 in y a = b x 2 + c x + d , we will get b + c + d = − 6 .
For the equation to be a parabola, a=1.
Hence, a + b + c + d = − 5 ⇒ ∣ a + b + c + d ∣ = 5