What is the extreme p(x)=12x^4+21x^3+ax^2-21x-9 ?

The extreme p(x)=12x^4+21x^3+ax^2-21x-9 is None

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extreme p(x)=12x^4+21x^3+ax^2-21x-9

extreme points

p (x) = 12 x^{4} + 21 x^{3} + a x^{2} - 21 x - 9

None

Solution steps

Find the critical points:

None

12 x^{4} + 21 x^{3} + a x^{2} - 21 x - 9

has no critical points, therefore there are no extreme points

None

2 x^{4} - 12 x^{2}

- 10 x^{2} + 180 x + 10

f (x) = x x - 3

f (x) = 2 sec (x)

7 (4 x^{2} - 36)^{3} + 8

What is the extreme p(x)=12x^4+21x^3+ax^2-21x-9 ?
The extreme p(x)=12x^4+21x^3+ax^2-21x-9 is None