Asked by sh
Find the intervals of increase and decrease for the following function
h(x)=(x^3)(x-1)^4
h'(x)=[(3x^2)(x-1)^4]+(4x^3)(x-1)^3
How do I find the zeros?
Thanks.
h(x)=(x^3)(x-1)^4
h'(x)=[(3x^2)(x-1)^4]+(4x^3)(x-1)^3
How do I find the zeros?
Thanks.
Answers
Answered by
Reiny
The zeros of the derivative or the zeros of the function ?
I will assume you want the zeros of the derivative.
factor it first
= x^2(x-1)^3[4x + 3(x-1)]
= x^2(x-1)^3(7x-3)
we set this equal to zero to get
x = 0, x = 1 and x = 3/7
remember that the function increases when the first derivative is positive, and decreases when that derivative is negative.
Can you take it from here ?
I will assume you want the zeros of the derivative.
factor it first
= x^2(x-1)^3[4x + 3(x-1)]
= x^2(x-1)^3(7x-3)
we set this equal to zero to get
x = 0, x = 1 and x = 3/7
remember that the function increases when the first derivative is positive, and decreases when that derivative is negative.
Can you take it from here ?
Answered by
sh
Do you factor it by taking the lowest ^ ? I can continue from there, thanks! :)
Answered by
sh
The answer key indicated that 0 is not a zero.
Answered by
Reiny
Your question was,
"how do I find the zeros"
x = 0 is a solution to h(x) = 0 and h'(x) = 0
"how do I find the zeros"
x = 0 is a solution to h(x) = 0 and h'(x) = 0
Answered by
sh
If x=0, shouldn't it be used to determine the increase and decrease?