Question

Find the maximum and minimum values of 4xcube +9xsquare-12x+1

Answer 1

$\orange{ \sf \: 4x {}^{3} + 9x {}^{2} - 12 {x}^{} + 1}$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of

$\sf \: ax^{n} \: is \: \: nax^{n-1}.$

$\pink{ \sf = > 3\times 4x^{3-1}+2\times 9x^{2-1}-12x^{1-1}}$

$\sf \: \color{red}Multiply \: 3 \: times \: 4.$

$\pink{ \sf= > 12x^{3-1}+2\times 9x^{2-1}-12x^{1-1}}$

$\sf \: \color{red}Subtract \: 1 \: from \: 3..$

$\pink{ \sf = > 12x^{2}+2\times 9x^{2-1}-12x^{1-1}}$

$\sf \: \color{red}Multiply \: 2 \: times \: 9..$

$\pink{ \sf= > 12x^{2}+18x^{2-1}-12x^{1-1}}$

$\sf \: \color{red}Subtract \: 1 \: from \: 2..$

$\pink{ \sf= > 12x^{2}+18x^{1}-12x^{1-1}}$

$\sf \: \color{red}Subtract \: 1 \: from \: 1..$

$\pink{ \sf= > 12x^{2}+18x^{1}-12x^{0}}$

$\sf \: \color{red}For \: any \: term \: t, \: t^{1}$

$\sf \pink{= > 12x^{2}+18x-12x^{0}}$

$\sf\color{red}For \: any \: term \: t \: except \: 0, t^{0}=1.$