# Thread: Analyzing functions (calculus)

1. ## Analyzing functions (calculus)

Given a function, determine rising and falling intervals, concavity, critical points, points of inflection.

y = x(x-4)^3

In a function like this, would you expand everything and then try to do it or use the product rule? If you use product rule, the derivatives are a bit complex and I find it hard to see how to isolate x to determine when the functions are zero (to find certain points). Can someone show by example how to approach this?

2. Originally Posted by theowne
Given a function, determine rising and falling intervals, concavity, critical points, points of inflection.

y = x(x-4)^3

In a function like this, would you expand everything and then try to do it or use the product rule? If you use product rule, the derivatives are a bit complex and I find it hard to see how to isolate x to determine when the functions are zero (to find certain points). Can someone show by example how to approach this?
do the product rule. post your answer here and then we'll take it from there

3. y = x(x-4)^3
y'= x(3(x-4)^2) + (x-4)^3
y''= x{6(x-4)}+3(x-4)^2+3(x-4)^2=x{6(x-4)}+6(x-4)^2

Is that right?

4. Originally Posted by theowne
y = x(x-4)^3
y'= x(3(x-4)^2) + (x-4)^3
y''= x{6(x-4)}+3(x-4)^2+3(x-4)^2=x{6(x-4)}+6(x-4)^2

Is that right?
besides the fact that these are ugly forms, they are correct.

note that:

y' = 4(x - 1)(x - 4)^2

and

y'' = 12(x - 4)(x - 2)

it is not hard to equate either of those to zero and solve for x

5. Uh, could you show how you simplified to those forms please?

6. Originally Posted by theowne
Uh, could you show how you simplified to those forms please?
well, for the first one, when i did the product rule i ended up with:

y' = 3x(x - 4)^2 + (x - 4)^3

before moving on, i simplified this. factor out the (x - 4)^2 and you will get what i got. do a similar thing for the second. factor out te (x - 4) and simplify