The parabola (1 Viewer)

[mazza_728]

What am i doing wrong? i cant get this!

Locate max, min and inflexion points for the curve f(x)=2x^3+2x^2-2x-1

f'(x)= 6x^2+4x-2 and f"(x)=12x+4
Turning points at f'(x)=0
0=(3x-1)(2x+2)
x=1/3 or -1
y=0 or 0

Test (1/3,0) and (-1,0)
f"(1/3) > 0 min at (1/3,0)
f"(-1) < 0 max at (-1,0)

PPI at f"(x) = 0
0=12x+4
x=-1/3
y=-5/27

y intercept at -1

When i try and graph this .. it doesnt work

Please help if you can!

 

[Winston]

Originally posted by mazza_728


What am i doing wrong? i cant get this!

Locate max, min and inflexion points for the curve f(x)=2x^3+2x^2-2x-1

f'(x)= 6x^2+4x-2 and f"(x)=12x+4
Turning points at f'(x)=0
0=(3x-1)(2x+2)
x=1/3 or -1
y=0 or 0

Test (1/3,0) and (-1,0)
f"(1/3) > 0 min at (1/3,0)
f"(-1) < 0 max at (-1,0)

PPI at f"(x) = 0
0=12x+4
x=-1/3
y=-5/27

y intercept at -1

When i try and graph this .. it doesnt work

Please help if you can!

y=0 or 0 maybe that's a little wrong, when x = -1, y = 1, when x = 1/3 y = -37/27

that could very well be where your stuffing it up.

 

[iambored]

Originally posted by Winston
y=0 or 0 maybe that's a little wrong, when x = -1, y = 1, when x = 1/3 y = -37/27

that could very well be where your stuffing it up.

yeah i think you subbed the x values into y'. because that's the only way y would = 0 at both points. as you want a co ordinate you have to sub it into y=___ NOT y'=___

 

[ND]

You should have recognised that it's wrong as soon as you saw y=0, 0; how are you gonna have it pass through the same y-value without a stationary pt in between? (and there is obviously no cusp on this curve)

 

[mazza_728]

Thanks guys, i always make stupid mistakes and get so frustrated!