So a problem:
On which interval(s) does the function f(x)=x^4-4x^3+4x^2+6 increase?
Key word: increase, which means to take first derivative test
x^4-4x^3+4x^2+6
4x^3-12x^2+8x
4x(x^2-3x+2)
(x-2)(x-1)
x=0, 1, 2
(-infinity, 0) (0,1) (1,2) (2, infinity)
-1: 4(-1)^3-12(-1)^2+8(-1) = - ve
.5: 4(.5)^3-12(.5)^2+8(.5) = + ve
1.5: 4(1.5)^3-12(1.5)^2+8(1.5) = - ve
3: 4(3)^3-12(3)^2+8(3) = + ve
So 0
touble with:
I'm having trouble with problems from the non calculator section. I just don't know where to begin sometimes. But I'm mainly still having trouble with washers and the stuff like that.
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ReplyDeleteokay the blog thing messed up the So 02 thing is suppose to be 0 < x <1 and x > 2
ReplyDeleteThe formula for the area of washers is S (top) - (bottom)
ReplyDeleteSTEPS:1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate