Examining a Curve - # 2

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Jason76

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All of this is wrong. Please look thru for mistakes.

\(\displaystyle f(x) = x^{4} - 32x + 1\)

\(\displaystyle f'(x) = 4x^{3} - 32\)

\(\displaystyle f''(x) = 12x\)

Find Critical Points:

\(\displaystyle f'(x) = 4x^{3} - 32\)

\(\displaystyle 4x^{3} - 32 = 0\)

\(\displaystyle 4x^{3} = 32\)

\(\displaystyle x^{3} = 8\)

\(\displaystyle x = 2\) - \(\displaystyle 3rd\) root of \(\displaystyle 8\) is \(\displaystyle 2\)

What is the maximum and minimum?

\(\displaystyle f(x) = x^{4} - 32x + 1\)

\(\displaystyle f(2) = (2)^{2} - 32(2) + 1 = -59\)

Where is the interval increasing or decreasing?

\(\displaystyle f'(x) = 4x^{3} - 32\)

For \(\displaystyle x<2\)

\(\displaystyle f'(1) = 4(1)^{3} - 32 = -21\) Negative so decreasing

For \(\displaystyle 2>x\)

\(\displaystyle f'(3) = 4(x)^{3} - 32 = 4\) Positive so increasing

What is the inflection point?

\(\displaystyle f''(x) = 12x\)

\(\displaystyle 12x = 0\)

\(\displaystyle x = 0\)

\(\displaystyle f(x) = x^{4} - 32x + 1\)

\(\displaystyle f(0) = (0)^{4} - 32(0) + 1 = 1\)

Inflection Point \(\displaystyle (0,1)\)

Where is the curve concave up, or concave down?

For \(\displaystyle x<0\)

\(\displaystyle f''(x) = 12x\)

\(\displaystyle f''(-1) = 12(-1) = -12\) Negative so concave up

For \(\displaystyle x<0\)

\(\displaystyle f''(x) = 12x\)

\(\displaystyle f''(-1) = 12(-1) = -12\) Positive so concave down
 
What do you mean by "all of this is wrong"? On what basis have you determined this?
 
For starters, the second derivative is wrong. There are others, too.

Jason76, it seems like you do the problem (possibly very quickly due to the numerous careless errors we have found) and then once you see that your answer does not match the book's answer then you ask for help. I may be wrong in this thinking, but do you actually double check your work when you get an incorrect error?
 
:?: 2nd derivative is corrected. Any hints on other errors?

\(\displaystyle f(x) = x^{4} - 32x + 1\)

\(\displaystyle f'(x) = 4x^{3} - 32\)

\(\displaystyle f''(x) = 12x^{2}\)

Find Critical Points:

\(\displaystyle f'(x) = 4x^{3} - 32\)

\(\displaystyle 4x^{3} - 32 = 0\)

\(\displaystyle 4x^{3} = 32\)

\(\displaystyle x^{3} = 8\)

\(\displaystyle x = 2\) - \(\displaystyle 3rd\) root of \(\displaystyle 8\) is \(\displaystyle 2\)

What is the maximum and minimum?

\(\displaystyle f(x) = x^{4} - 32x + 1\)

\(\displaystyle f(2) = (2)^{2} - 32(2) + 1 = -59\)

Where is the interval increasing or decreasing?

\(\displaystyle f'(x) = 4x^{3} - 32\)

For \(\displaystyle x<2\)

\(\displaystyle f'(1) = 4(1)^{3} - 32 = -21\) Negative so decreasing

For \(\displaystyle 2>x\)

\(\displaystyle f'(3) = 4(x)^{3} - 32 = 4\) Positive so increasing

What is the inflection point?

\(\displaystyle f''(x) = 12x^{2}\)

\(\displaystyle 12x^{2} = 0\)

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

\(\displaystyle x = 0\)

\(\displaystyle f(x) = x^{4} - 32x + 1\)

\(\displaystyle f(0) = (0)^{4} - 32(0) + 1 = 1\)

Inflection Point \(\displaystyle (0,1)\)

Where is the curve concave up, or concave down?

For \(\displaystyle x<0\)

\(\displaystyle f''(x) = 12x^{2}\)

\(\displaystyle f''(-1) = 12(-1)^{2} = 12\) Negative so concave up

For \(\displaystyle 0 < x\)

\(\displaystyle f''(x) = 12x\)

\(\displaystyle f''(1) = 12(1)^{2} = 12\) Positive so concave up
 
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Jason76 said:
:?: 2nd derivative is corrected. Any hints on other errors?

\(\displaystyle f(x) = x^{4} - 32x + 1\)

\(\displaystyle f'(x) = 4x^{3} - 32\)

\(\displaystyle f''(x) = 12x^{2}\)



Find Critical Points:\(\displaystyle \ \ \ \) <---- No, you are finding a "critical number."
---------------------------


\(\displaystyle \ \ \ \) <---- No, you are finding a "critical number."

\(\displaystyle f'(x) = 4x^{3} - 32\)

\(\displaystyle 4x^{3} - 32 = 0\)

\(\displaystyle 4x^{3} = 32\)

\(\displaystyle x^{3} = 8\)

\(\displaystyle x = 2\) - \(\displaystyle 3rd\) root of \(\displaystyle 8\) is \(\displaystyle 2\)

What is the maximum and minimum?
-------------------------------------------------


\(\displaystyle f(x) = x^{4} - 32x + 1\)

\(\displaystyle f(2) = (2)^{2} - 32(2) + 1 = -59 \ \ \ \) Wrong. It's supposed to be \(\displaystyle \ (2)^4 - 32(2) + 1 = -47\)

And you haven't shown whether that is a maximum or minimum value.



Where is the interval increasing or decreasing?
-------------------------------------------------------------


\(\displaystyle f'(x) = 4x^{3} - 32\)

For \(\displaystyle x<2\)

\(\displaystyle f'(1) = 4(1)^{3} - 32 = -21 \ \ \ \) No, that equals -28. \(\displaystyle \ \ \ \ \)Negative so decreasing

For \(\displaystyle 2>x \ \ \ \)No. That means the same thing as "x < 2."[/b]

\(\displaystyle f'(3) = 4(x)^{3} - 32 = 4 \ \ \ \)No. That should be \(\displaystyle \ 4(3)^3 - 32 = 4(27) - 32 = 108 - 32 = 76. \ \ \ \ \) Positive so increasing

What is the inflection point?
---------------------------------------


\(\displaystyle f''(x) = 12x^{2}\)

\(\displaystyle 12x^{2} = 0\)

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

\(\displaystyle x = 0\)

\(\displaystyle f(x) = x^{4} - 32x + 1\)

\(\displaystyle f(0) = (0)^{4} - 32(0) + 1 = 1\)

Inflection Point \(\displaystyle (0,1) \ \ \ \)No. Look below.

Where is the curve concave up, or concave down?
-----------------------------------------------------------------


For \(\displaystyle x<0\)

\(\displaystyle f''(x) = 12x^{2}\)

\(\displaystyle f''(-1) = 12(-1)^{2} = 12\) Negative \(\displaystyle \ \ \ \)<--- No, it's not negative, it's positive. so concave up

For \(\displaystyle 0 < x \ \ \ \)

\(\displaystyle f''(x) = 12x\)

\(\displaystyle f''(1) = 12(1)^{2} = 12\) Positive so concave up \(\displaystyle \ \ \ \ \)x = 0 was a candidate for being the x-value

for a point of inflection, but the graph is concave up on either side of x = 0. Therefore, there is no point

of inflection at x = 0. And as x = 0 was the only candidate that came from f''(x) = 0, then there are no

points of inflection.
Click to expand...
.
 
Inflection Point

\(\displaystyle 12x^{2} - 64x = 0\)

\(\displaystyle 4x(3x - 16) = 0\)

\(\displaystyle 4x = 0\)

\(\displaystyle x = 0\)

\(\displaystyle 3x - 16 = 0\)

\(\displaystyle 3x = 16\)

\(\displaystyle x = \dfrac{3}{16}\)

Inflection Point - # 1

\(\displaystyle f''(0) = 12(0)^{2} - 64 = -64\)

Inflection Point - # 2

\(\displaystyle f''(0) = 12(\dfrac{3}{16})^{2} - 64 = \)
 
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Jason76 said:
Inflection Point

\(\displaystyle 12x^{2} - 64x = 0\)

\(\displaystyle 4x(3x - 16) = 0\)

\(\displaystyle 4x = 0\)

\(\displaystyle x = 0\)

\(\displaystyle 3x - 16 = 0\)

\(\displaystyle 3x = 16\)

\(\displaystyle x = \dfrac{3}{16}\)

Inflection Point - # 1

\(\displaystyle f''(0) = 12(0)^{2} - 64 = -64\)

Inflection Point - # 2

\(\displaystyle f''(0) = 12(\dfrac{3}{16})^{2} - 64 = \)
Click to expand...

Your whole post (the quote box) is wrong. Your second derivative is wrong. In the previous post of mine,

I already spelled out that the function in question has no points of inflection.
 
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