What is the extreme y=2x^3+ax^2-12x-4,x=1 ?

The extreme y=2x^3+ax^2-12x-4,x=1 is {a>-6: Minimum(1,a-14),a<-6: Maximum(1,a-14)}

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extreme y=2x^3+ax^2-12x-4,x=1

extreme points

y = 2 x^{3} + a x^{2} - 12 x - 4, x = 1

{a > - 6 : Minimum (1, a - 14), a < - 6 : Maximum (1, a - 14)}

Solution steps

Find the critical points:

x = 1

f^{''} (x) = 12 x + 2 a

Check the sign of

f^{''} (x) = 12 x + 2 a

at each critical point

Check critical point

x = 1 : {a > - 6 : Minimum (1, a - 14), a < - 6 : Maximum (1, a - 14)}

{a > - 6 : Minimum (1, a - 14), a < - 6 : Maximum (1, a - 14)}

4 a p + 50 p - 9 p^{2} - \frac{1}{10} a^{2} p - 110

2 x^{3} + 5 x^{2} + 5 y^{3} - 5 y^{2} + 10

f (x, y) = x^{3} + 3 x^{2} + 6 x y + y^{2}

\frac{( 2 x + 1 )}{( e ^{x^{2}} )}

f (y) = 3 y^{2} - 3 x^{2} - 6 y

What is the extreme y=2x^3+ax^2-12x-4,x=1 ?
The extreme y=2x^3+ax^2-12x-4,x=1 is {a>-6: Minimum(1,a-14),a<-6: Maximum(1,a-14)}