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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)^3Quote:
Originally Posted by UMStudent
=> 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
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]
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 rootQuote:
Originally Posted by UMStudent
f(x) = (x^4)(x-1)^3Quote:
Originally Posted by UMStudent
=> 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
Ah alright thanks I see it now, sometimes its hard to see things when they are written on the computer i think.
that's trueQuote:
Originally Posted by UMStudent
sometimes writing stuff out on paper makes it clearer