1. ## Critical numbers

need to find critical numbers of f(x) = x^4(x-1)^3

so I took the derivative using the product rule

(x-1)^3*4x^3 + x^4*3(x-1)^2

worked that out to

4x^3(x^3-3x^2+3x-1) + 3x^4(x^2-2x+1)

then I'm kinda lost its obvious one critical point is at 0 but there should be at least 1 more 2 if there is an inflection point. Please help me out I'm really lost on this stuff and these problems are due tomorrow.

2. Originally Posted by UMStudent
need to find critical numbers of f(x) = x^4(x-1)^3

so I took the derivative using the product rule

(x-1)^3*4x^3 + x^4*3(x-1)^2

worked that out to

4x^3(x^3-3x^2+3x-1) + 3x^4(x^2-2x+1)

then I'm kinda lost its obvious one critical point is at 0 but there should be at least 1 more 2 if there is an inflection point. Please help me out I'm really lost on this stuff and these problems are due tomorrow.
f(x) = (x^4)(x-1)^3
=> f ' (x) = (4x^3)(x - 1)^3 + 3(x^4)(x - 1)^2
=> f ' (x) = (x^3)(x - 1)^2 * [4(x - 1) + 3x]
=> f ' (x) = (x^3)(x - 1)^2 * [7x - 4]

for critical numbers we have f ' (x) = 0
=> (x^3)(x - 1)^2 * [7x - 4] = 0
=> (x^3) = 0
or (x - 1)^2 = 0
or 7x - 4 = 0

=> x = 0
or x = 1
or x = 4/7

3. I follow your first step there thats what I had for the first derivative but I don't see how you go from the derivative to the simplified function and then from that one to the further simplified function.

f(x) = (x^4)(x-1)^3
=> f ' (x) = (4x^3)(x - 1)^3 + 3(x^4)(x - 1)^2
=> f ' (x) = (x^3)(x - 1)^2 * [4(x - 1) + 3x]
=> f ' (x) = (x^3)(x - 1)^2 * [7x - 4]

4. Originally Posted by UMStudent
its obvious one critical point is at 0 but there should be at least 1 more 2 if there is an inflection point.
note that when expanding f(x) you get an x^7 graph, that means by rite you should have 7 zeroes in f(x) and 6 zeroes in f ' (x). how come we only have three? some are repeated, the x^3 yields a triple root, the (x - 1)^2 yields a double root and the 7x - 4 yields one root
Originally Posted by UMStudent
I follow your first step there thats what I had for the first derivative but I don't see how you go from the derivative to the simplified function and then from that one to the further simplified function.

f(x) = (x^4)(x-1)^3
=> f ' (x) = (4x^3)(x - 1)^3 + 3(x^4)(x - 1)^2
=> f ' (x) = (x^3)(x - 1)^2 * [4(x - 1) + 3x]
=> f ' (x) = (x^3)(x - 1)^2 * [7x - 4]
f(x) = (x^4)(x-1)^3
=> f ' (x) = (4x^3)(x - 1)^3 + 3(x^4)(x - 1)^2
=> f ' (x) = (x^3)(x - 1)^2 * [4(x - 1) + 3x] ............here i factorized, (x^3) and (x - 1) are common to both terms
=> f ' (x) = (x^3)(x - 1)^2 * [7x - 4] .....................here i just simplified what's in the square brackets

5. Ah alright thanks I see it now, sometimes its hard to see things when they are written on the computer i think.

6. Originally Posted by UMStudent
Ah alright thanks I see it now, sometimes its hard to see things when they are written on the computer i think.
that's true

sometimes writing stuff out on paper makes it clearer