Critical Points of f(x): -2x^3 + 39x^2 - 180x + 1

[smith5029]

hi guys I'm stuck on this, i keep thinking i have the right answeres but i can't get them.
the function is f(x): -2x^3 + 39x^2 - 180x + 1
i need to list all of the critical points, and hten indicate where it is increasing and where it is decreasing.

I set the derivative -6x^2 + 78x - 180 = 0 and then solved for x. this is correct right. by doing so i ended up with critical points at -2 and 15. Then made a number line and figured that the slope before -2 was neg, b/w -2 and 15 it was pos, and after 15 it was neg. so my interval of increasing was (-2,15) and the interval of decreasing was (-inf, -2) U (15,inf) where am i going wrong - PLEASE HELP ME

 

[dextercioby]

Did u compute the second derivative,too...?To find the nature of those 2 points,u must evaluate the second derivative in those points.

Daniel.

 

[mathman]

You made a mistake in solving the quadratic. The roots are 3 and 15.

 

[smith5029]

Yes i determined the second derivative as -12x + 78. Then using the second derivative rule plugged in 15, and -2 telling me that -2 was a local min and 15 was a local max. DO you know if 15, -2 are the only critical points in my problem?

 

[smith5029]

i don't see how u got 3 and 15 ??

 

[dextercioby]

The roots are 3 & 10.

Daniel.

 

[cronxeh]

-6x^2 + 78x - 180 = 0

x^2 -13x + 30 = 0
x=(13+/-sqrt(169-4*1*30))/2
x=(13+/-7)/2

x=20/2 = 10
x=6/2= 3