Calculus help (1 Viewer)

[astroman]

 

[seventhroot]

these questions pop up in the HSC a lot

let f(x) = 3x^2 - x^3

diff the function

dy/dx = 6x - 3x^2

solve dy/dx = 0 for stat points

hence x = 0, 2. for the y values sub this into f(x) {NOT dy/dx}

now for nature

d^2y/dx^2 = 6(1 - x)

subbing in x = 0 into d^2y/dx^2 we get a +ve value so it is a local min and when we sub in x = 2 we get a -ve value hence it is a local max

 

[astroman]

where did u get d^2y/dx^2 = 6(1 - x) from?

 

[Menomaths]

where did u get d^2y/dx^2 = 6(1 - x) from?

dy/dx = 6x - 3x^2

differentiate this again to get d^2y/dx^2 = 6-6x = 6(1-x)

 

[astroman]

oh ok

 

[astroman]

answer says (0,0) is a min and (2.4) is a max

how?

 

[seventhroot]

answer says (0,0) is a min and (2.4) is a max

how?

what are you confused about? how they got the y values?

 

[astroman]

yes, do i have to find x and y intercepts?

 

[seventhroot]

yes, do i have to find x and y intercepts?

so what they did was sub x = 0 into f(x) which is 0; hence the coordinates are (0, 0) likewise with x = 2

yes if you are being required to sketch the graph

 

[astroman]

got it, thanks

 

[seventhroot]

nws