Critical Number Problem - # 3

[Jason76]

\(\displaystyle f(x) = (x^{2} - 1)^{3}\) on interval \(\displaystyle [-1,4]\)

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

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

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

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

\(\displaystyle x = \pm \sqrt{1}\)

b. \(\displaystyle 2x = 0\)

\(\displaystyle x = 0\)

 

[Deleted member 4993]

\(\displaystyle f(x) = (x^{2} - 1)^{3}\) on interval \(\displaystyle [-1,4]\)

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

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

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

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

\(\displaystyle x = \pm \sqrt{1}\)

b. \(\displaystyle 2x = 0\)

\(\displaystyle x = 0\)

.

 

[Jason76]

.

corrected

\(\displaystyle f(x) = (x^{2} - 1)^{3}\) on interval \(\displaystyle [-1,4]\)

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

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

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

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

\(\displaystyle x = \pm \sqrt{1}\)

b. \(\displaystyle 6x = 0\)

\(\displaystyle x = 0\)

 

[Deleted member 4993]

corrected

\(\displaystyle f(x) = (x^{2} - 1)^{3}\) on interval \(\displaystyle [-1,4]\)

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

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

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

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

\(\displaystyle x = \pm \sqrt{1}\) = ± 1

b. \(\displaystyle 6x = 0\)

\(\displaystyle x = 0\)

.

 

[Jason76]

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The \(\displaystyle -1\) answer is right for the min, but \(\displaystyle 1\) doesn't work for the max

 

[Deleted member 4993]

The \(\displaystyle -1\) answer is right for the min, but \(\displaystyle 1\) doesn't work for the max

Global max. is at x=4

 

[Jason76]

Global max. is at x=4

Computer still says wrong for the max

 

[Deleted member 4993]

The \(\displaystyle -1\) answer is right for the min, but \(\displaystyle 1\) doesn't work for the max

The minimum should be at x = 0

f(-1) = 0

f(0) = -1

 

[Jason76]

\(\displaystyle f(x) = (x^{2} - 1)^{3}\) on interval \(\displaystyle [-1,4]\)

The computer says \(\displaystyle -1\) is the min (I guessed it), but my calculations don't seem to get it.

\(\displaystyle f(-1) = ((-1)^{2} - 1)^{3} = 0\)

\(\displaystyle f(0) = ((0)^{2} - 1)^{3} = 1\) Is this \(\displaystyle -1\)

\(\displaystyle f(1) = ((1)^{2} - 1)^{3} = 0\)


\(\displaystyle f(4) = ((4)^{2} - 1)^{3} = 3375\) Max

 

[lookagain]

\(\displaystyle f(x) = (x^{2} - 1)^{3}\) on interval \(\displaystyle [-1,4]\)

The computer says \(\displaystyle -1\) is the min (I guessed it), but my calculations don't seem to get it.

\(\displaystyle f(-1) = ((-1)^{2} - 1)^{3} = 0\)

\(\displaystyle f(0) = ((0)^{2} - 1)^{3} = 1\) Is this -1 \(\displaystyle \ \ \ \ \)That does equal -1.

\(\displaystyle f(1) = ((1)^{2} - 1)^{3} = 0\) \(\displaystyle f(4) = ((4)^{2} - 1)^{3} = 3375\) Max

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