Question

Ex 2.11
20) A curve y''=12x-24 and a stationary point at (1,4). evaluate y when x=2.

Answer 1

`y''=12x-24\\ y'=\int 12x-24\, dx\\ y'=6x^2-24x+C\\\\ 0=6\cdot1^2-24\cdot1+C\\ 0=6-24+C\\ C=18\\ y'=6x^2-24x+18\\\\ y=\int 6x^2-24x+18\, dx\\ y=2x^3-12x^2+18x+C\\\\ 4=2\cdot1^3-12\cdot1^2+18\cdot1+C\\ 4=2-12+18+C\\ C=-4\\\\ 2x^3-12x^2+18x-4`


`y(2)=2\cdot2^3-12\cdot2^2+18\cdot2-4\\ y(2)=16-48+36-4\\ \boxed{y(2)=0}`

Answer 2

So, dy/dx=0 at the point (1, 4) - that is where x=1 and y=4.


`\int { 12x-24dx } \\ \\ =\frac { 12{ x }^( 2 ) }{ 2 } -24x+C\\ \\ =6{ x }^( 2 )-24x+C`


`\\ \\ \therefore \quad { f }^( ' )\left( x \right) =6{ x }^( 2 )-24x+C`

But when x=1, f'(x)=0, therefore:


`0=6-24+C\\ \\ 0=-18+C\\ \\ \therefore \quad C=18`


`\\ \\ \therefore \quad { f }^( ' )\left( x \right) =6{ x }^( 2 )-24x+18`

Now:


`\int { 6{ x }^( 2 ) } -24x+18dx\\ \\ =\frac { 6{ x }^( 3 ) }{ 3 } -\frac { 24{ x }^( 2 ) }{ 2 } +18x+C`


`=2{ x }^( 3 )-12{ x }^( 2 )+18x+C\\ \\ \therefore \quad f\left( x \right) =2{ x }^( 3 )-12{ x }^( 2 )+18x+C`

Now when x=1, y=4:


`4=2-12+18+C\\ \\ 4=8+C\\ \\ C=4-8\\ \\ C=-4`


`\\ \\ \therefore \quad f\left( x \right) =2{ x }^( 3 )-12{ x }^( 2 )+18x-4`

Now when x=2,


`f\left( x \right) =2\cdot { 2 }^( 3 )-12\cdot { 2 }^( 2 )+18\cdot 2-4\\ \\ =16-48+36-4\\ \\ =0`

So when x=2, y=0.