Prove that the function $y = 3 x^{4} - 4 x^{3} + 6 x^{2} + a x + b$ has only one point of extremum for any a and b.

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Solution

$y = {3 x}^{4} - {4 x}^{3} + {6 x}^{2} + a x + b$
$y^{1} = {12 x}^{3} - {12 x}^{2} + 12 x + a$
$y^{11} = {36 x}^{2} - 24 x + 12$
from which it's cleared that $y$ function has atremom irrespective of $a$ & $b$ .

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Similar questions

a, b \in R

f (x) = 3 x^{4} - 4 x^{3} + 6 x^{2} + a x + b

\in R

f (x) = 3 x^{4} - 4 x^{3} + 6 x^{2} + a x + b

a, b ϵ R

f (x) = 3 x^{4} - 4 x^{3} + 6 x^{2} + a x + b

a

b

y = a {log}_{e} x + b x^{2} + x

x_{1} = 1

x_{2} = 2

f (x) = 3 x^{4} - 4 x^{3}

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