Max and Min Problem

[Jason76]

\(\displaystyle f(x)= (x^{2} - 1)^{3}\)

\(\displaystyle f'(x) = 3 (u)^{2} du\)

\(\displaystyle f'(x) = 3 (u)^{2} 2x\)

\(\displaystyle f'(x) = 6x (u)^{2}\)

\(\displaystyle f'(x) = 6x (x^{2} - 1)^{2}\)

\(\displaystyle 6x (x^{2} - 1)^{2} = 0\)

\(\displaystyle 6x = 0\)

\(\displaystyle x = 0\)

AND

\(\displaystyle (x^{2} - 1)(x^{2} - 1) = 0\)

\(\displaystyle x = 1\)

Find 2nd derivative for max min

\(\displaystyle f''(x) = \dfrac{d^{2}}{dx} (x^{2} - 1)^{2} \)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][2 (u)^{2} du]\)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][2 (u)^{2} 2x]\)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][(u)^{2} 4x]\)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][(x^{2} - 1)^{2} 4x]\)

 

[stapel]

\(\displaystyle f(x)= (x^{2} - 1)^{3}\)

What were the instructions? What are you supposed to be doing with this function?

\(\displaystyle f'(x) = 6x (x^{2} - 1)^{2}\)

Find 2nd derivative for max min

The second derivative of f(x) is the first derivative of f'(x), so differentiate the above function.

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][(x^{2} - 1)^{2} 4x]\)

The is ambiguous. What, exactly, is your question?

 

[HallsofIvy]

\(\displaystyle f(x)= (x^{2} - 1)^{3}\)

\(\displaystyle f'(x) = 3 (u)^{2} du\)

The "du" here is incorrect. It should be \(\displaystyle f'(x)= 3(u)^2\frac{du}{dx}\).
And you should say that you are letting \(\displaystyle u= x^2- 1\) before writing it.

\(\displaystyle f'(x) = 3 (u)^{2} 2x\)

\(\displaystyle f'(x) = 6x (u)^{2}\)

\(\displaystyle f'(x) = 6x (x^{2} - 1)^{2}\)

Yes, this is the derivative of f.

\(\displaystyle 6x (x^{2} - 1)^{2} = 0\)

Assuming you want to find max and min values of f, yes, you set the derivative of f to 0.
But don't just put your objective in the title! State it at the beginning of your post.

\(\displaystyle 6x = 0\)

\(\displaystyle x = 0\)

AND

\(\displaystyle (x^{2} - 1)(x^{2} - 1) = 0\)

\(\displaystyle x = 1\)

You are missing a value of x.

Find 2nd derivative for max min

\(\displaystyle f''(x) = \dfrac{d^{2}}{dx} (x^{2} - 1)^{2} \)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][2 (u)^{2} du]\)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][2 (u)^{2} 2x]\)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][(u)^{2} 4x]\)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][(x^{2} - 1)^{2} 4x]\)

Why in the world do you start all over again to find the second derivative? You already knew that
\(\displaystyle f'= 6x(x^2- 1)^2\) so \(\displaystyle f''= 6(x^2- 1)^2+ (6x)(2(x^2- 1)(2x))= 6(2x-1)^2+ 24x(x^2-1)\)
Now, what do you do with that?


Personally, I think I would have used the "first derivative test". \(\displaystyle (x^2- 1)^2\) is not negative for any x so the sign of the derivative depends entirely on "6x" which is positive for x> 0 and negative for x< 0. As we go from left to right (x increasing) the graph is going down to (0, -1) then up again so that is a miniumum. But f' is positive on both sides of x= 1 so that gives neither a maximum nor a minimum. And, as I said before, you are missing one critical point.

 

[Jason76]

You would respectively plug in each critical point (into the 2nd derivative) to see if it's a max or min. But first, we have to make sure the 2nd derivative is correct.

 

[Deleted member 4993]

\(\displaystyle f(x)= (x^{2} - 1)^{3}\)

\(\displaystyle f'(x) = 3 (u)^{2} du\)

\(\displaystyle f'(x) = 3 (u)^{2} 2x\)

\(\displaystyle f'(x) = 6x (u)^{2}\)

\(\displaystyle f'(x) = 6x (x^{2} - 1)^{2}\)

\(\displaystyle 6x (x^{2} - 1)^{2} = 0\)

\(\displaystyle 6x = 0\)

\(\displaystyle x = 0\)

AND

\(\displaystyle (x^{2} - 1)(x^{2} - 1) = 0\)

\(\displaystyle x = 1\)

Find 2nd derivative for max min

\(\displaystyle f''(x) = \dfrac{d^{2}}{dx} (x^{2} - 1)^{2} \) ...Incorrect.. that should be \(\displaystyle f''(x) = \dfrac{d^{2}}{dx} (x^{2} - 1)^{3} \)
\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][2 (u)^{2} du]\)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][2 (u)^{2} 2x]\)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][(u)^{2} 4x]\)

\(\displaystyle f''(x) = [(x^{2} - 1)^{2}][6] + [6x][(x^{2} - 1)^{2} 4x]\)

.

 

[HallsofIvy]

The first derivative is, as you said, \(\displaystyle y'= 6x(x^2- 1)^2\) which is 0 at x= 0, x= 1, and x= -1.

The second derivtive is, of course, \(\displaystyle y''= 6(x^2- 1)^2+ (6x)(2)(x^2- 1)(2x))\). You can substitute 0, 1, and -1 directly into that: When x= 0, the first term is 6(1) and the second is 0 so the second derivative is positive and there is a relative minimum at x= 0. When x= 1 or -1, the \(\displaystyle x^2- 1\) terms are 0 so y''= 0. The "second derivative test" does NOT tell us anything.

But I would be inclined to use the "first derivative test" anyway. It is easy to see that \(\displaystyle y'= 6x(x-1)^2(x+ 1)^2\). If x< -1, all all three of x, x-1 and x+ 1 are negative. Since the squares are never negative, we have "negative*positive*positive" which is negative. Since the first derivative is negative, y is decreasing. Between -1 and 0, x+ 1 has changed sign to positive but because it is squared we still have "negative*positive*positive" and y is still decreasing. x= -1 is neither a maximum nor a minimum- it is a "inflection point" ("inflexion point" for our British friends). If x is between 0 and 1, the "x" term changes sign so y' is now positive: we have "positive*positive*positive". y is now increasing. Since we went down to x= 0 and are not going up, x= 0 is a minimum. Finally, for x> 1, while x- 1 has changed sign, since it is squared we still have "positive*positive*positive" so y' is positive and y is still increasing. Again, there neither a maximum nor a minimum at x= 1- it is again an inflection point.