hey guys, bad news, little error in second derivative ( -144x, not -148x)

here's what i got:

f(x)=2x^4 - 24x^3 + 96x^2 - 128x

f'(x)=8x^3 - 72x^2 + 192x - 128

f"(x)=24x^2 - 144x + 192

now to find min/max:

set f'(x)=0 and solve for x

so we get: 0=8x^3 - 72x^2 + 192x - 128

divide both sides by 8: 0=x^3 - 9x^2 + 24x - 16

now rewrite with brackets: 0=x^3 - (9x^2 - 24x + 16)

the last step is a trick i learned back in first year

now factor: 0=x^3 - (3x - 4)^2

this also takes some practice to see, although the quadratic equation will give you the roots

now rewrite: x^3=(3x-4)^2

x=1,x=4 only solutions use maple v7 (software, its great), but i got it by just plugging in numbers

increasing/decreasing intervals:

for all x<1, f'(x)<0 so decreasing

for all 1<x<4, f'(x)>0 so increasing

for all x>4, f'(x)>0 so increasing

thus x=4 not min/max point and x=1 is local and absolute min

plug x=1 into original: f(1)=-54

so min point is (1,-54) and x<1 is decreasing, x>1 is increasing

now to find points of inflection:

set f"(x)=0 and solve for x

so we get 0=24x^2 - 144x + 192

divide both side by 24: 0=x^2 - 6x + 8

factor: 0= (x-2)(x-4)

so x=2,x=4 agrees with software

find points of inflection by subbing into f(x):

f(2)=-32,f(4)=0, so points are (2,-32) and (4,0)

concavity:

for all x<2, f"(x)>0, so concave up

for all 2<x<4, f"(x)<0, so concave down

for all x>4, f"(x)>0, so concave up

also the graph passes through (0,0), know this, cause you can divide out the x from the 2x term

i know this is probably late for you but it was fun anyway. you have no idea how much money one can make by doing assignments for first years, i do, been doing so for 5 yrs.

cheers, pete