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I'm currently doing AP Calculus AB homework, and need some help with this problem. It sucks to forget. I know how to get the first and second derivative. I know that critical numbers can be found using the first derivative, and that points of inflection and concavity intervals can be found by using the second derivative, but I honestly forgot how to do it.

Analyze and sketch the graph of $f(x) = x^4 - 12x^3 + 48x^2 - 64x$

Edit: Solved!

  • First Derivative: $4x^3 - 36x^2 + 96x - 64$
  • Second Derivative: $12x^2 - 72x + 96$
  • X-Intercepts: $(0,0)$ $(4,0)$
  • Y-Intercepts: $(0,0)$
  • Critical Numbers: $X = 1$, $X = 4$
  • Points of Inflection: $X = 2$, $X = 4$
  • Increasing Intervals: $(1, Infinity)$
  • Decreasing Intervals: $(-Infinity, 1)$
  • Concavity Intervals: $(-Infinity, 2)$ $(2,4)$ $(4, Infinity)$
  • Relative Minimums: $X=1$
  • Relative Maximums: $X=2$
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This should bring you forward:

  • X- Intercepts: f(x) =0
  • Y- Intercepts: f(0)
  • critical numbers: f'(x) = 0
  • Points of inflections: f''(x) = 0 and f'''(x) $\neq$ 0
  • Increasing intervals: f'(x) > 0
  • Decreasing intervals: f'(x) < 0
  • Concavity upward: f''(x) > 0
  • Concavity downward f''(x) < 0
  • Relative Minimums: f'(x) = 0 (see above: critical numbers) and f''(x)>0
  • Relative Maximums: f'(x) = 0 (see above: critical numbers) and f''(x)<0

However, you should - at least once - really understand those criteria (e.g. graphically) rather than just apply them.

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  • $\begingroup$ I'm not really too sure on how to find the second X-intercept (one of them being 0), the Y-intercept is definitely 0 due to every number having an X variable. I'm confused when it comes to setting the first derivative equal to 0, I've factored out 4 and now i'm left with $4(x^3 - 9x^2 + 14x - 16) = 0$ $\endgroup$ – Khon Duong Jan 9 '14 at 10:15
  • $\begingroup$ For the second X-intercept: factor out x and then try out the divisors of the number without x (which will be 64). So one of the number 1,2,4,8,16,32,64 or their negative values should do it. Start with the smallest ones. Are you not allowed to use a calculator? $\endgroup$ – Bernd Jan 9 '14 at 10:27
  • $\begingroup$ Not for this particular question, no. The main focus of this problem is to be able to completely or accurately sketch a graph based on the characteristics that you get from doing all of those. $\endgroup$ – Khon Duong Jan 9 '14 at 10:35
  • $\begingroup$ For the first derivative: You factored out 4. That's correct. However it's 24, not 14. So for a product to be zero, at least one factor has to be zero. That means your 2nd factor must be 0 (since 4 is not). Again here: try divisors of the absolute number (without x: 16). Start with the smallest: 1,2, .. $\endgroup$ – Bernd Jan 9 '14 at 10:37
  • $\begingroup$ Just for clarification, we're solving for the critical number right? And the problem is still $4(x^3 - 9x^2 + 14x -16)$ and you want me to factor it a second time? $\endgroup$ – Khon Duong Jan 9 '14 at 10:47

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