Examining a Curve - # 3

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Jason76

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All of this is wrong. Please look over it.

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle f''(x) = 24x + 18\)

Find Critical Numbers:

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x + 3x - 9) = 0\)

\(\displaystyle \dfrac{-3 \pm \sqrt{3^{2} - 4(2)(-9)}}{2(2)}\)

\(\displaystyle \dfrac{-3 \pm \sqrt{81}}{4}\)

\(\displaystyle \dfrac{-3 \pm 9}{4}\)

\(\displaystyle \dfrac{-3 + 9}{4} = \dfrac{6}{4} = \dfrac{3}{4} \)

\(\displaystyle \dfrac{-3 - 9}{4} = \dfrac{-12}{4} = -3 \)

Critical Numbers: \(\displaystyle \dfrac{3}{4}, -3\)

Find Max Min

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f(\dfrac{3}{4}) = 4(\dfrac{3}{4})^{3} + 9(\dfrac{3}{4})^{2} - 54(\dfrac{3}{4}) + 8 = \)

\(\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 215 \)

Where does the interval increase or decrease?

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

For \(\displaystyle -3 < x <\dfrac{3}{4}\)

\(\displaystyle f'(-1) = 12(-1)^{2} + 18(-1) - 54 = \)

\(\displaystyle f'(-1) = 12 - 18 - 54 = -60\) Negaive so decreasing

For \(\displaystyle x < -3\)

\(\displaystyle f'(-4) = 12(-4)^{2} + 18(-4) - 54 = \)

\(\displaystyle f'(-4) = 192 -72 - 54 = 66 \)
 
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\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle f''(x) = 24x + 18\)

Find Critical Numbers:

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x + 3x - 9) = 0\)

\(\displaystyle \dfrac{-3 \pm \sqrt{3^{2} - 4(2)(-9)}}{2(2)}\)

\(\displaystyle \dfrac{-3 \pm \sqrt{81}}{4}\)

\(\displaystyle \dfrac{-3 \pm 9}{4}\)

\(\displaystyle \dfrac{-3 + 9}{4} = \dfrac{6}{4} = \dfrac{3}{2} \)

\(\displaystyle \dfrac{-3 - 9}{4} = \dfrac{-12}{4} = -3 \)

Critical Numbers: \(\displaystyle \dfrac{3}{2}, -3\)

Find Max Min

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 = -\dfrac{370}{8}\)

\(\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 215 \)

Where does the interval increase or decrease?

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

For \(\displaystyle x < -3\)

\(\displaystyle f'(-4) = 12(-4)^{2} + 18(-4) - 54 = \)

\(\displaystyle f'(-4) = 192 -72 - 54 = 66 \) Positive so increasing

For \(\displaystyle -3 < x <\dfrac{3}{2}\)

\(\displaystyle f'(-1) = 12(-1)^{2} + 18(-1) - 54 = \)

\(\displaystyle f'(-1) = 12 - 18 - 54 = -60\) Negative so decreasing

For \(\displaystyle \dfrac{3}{4} < x\)

\(\displaystyle f'(1) = 12(1)^{2} + 18(1) - 54 = - 24 \) Negative so decreasing

What is the infection point?

\(\displaystyle f''(x) = 24x + 18\)

\(\displaystyle 24x + 18 = 0\)

\(\displaystyle 24x = -18\)

\(\displaystyle x = -\dfrac{4}{3}\)

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{4}{3})^{3} + 9(-\dfrac{4}{3})^{2} - 54(-\dfrac{4}{3}) + 8 \)

\(\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{64}{67}) + 9(\dfrac{16}{9}) - 54(-\dfrac{4}{3}) + 8 \)

\(\displaystyle f(-\dfrac{4}{3}) = -\dfrac{256}{67}) + 16 + \dfrac{216}{3}) + 8 \) :confused:
 
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Jason76 said:
\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\) =======> Correct

\(\displaystyle f''(x) = 24x + 18\) =======> Correct

Find Critical Numbers:

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\) =======> Correct

\(\displaystyle 6(2x + 3x - 9) = 0\) =======> Correct

\(\displaystyle \dfrac{-3 \pm \sqrt{3^{2} - 4(2)(-9)}}{2(2)}\) =======> Correct

\(\displaystyle \dfrac{-3 \pm \sqrt{81}}{4}\) =======> Correct

\(\displaystyle \dfrac{-3 \pm 9}{4}\) =======> Correct

\(\displaystyle \dfrac{-3 + 9}{4} = \dfrac{6}{4} = \dfrac{3}{2} \) =======> Correct

\(\displaystyle \dfrac{-3 - 9}{4} = \dfrac{-12}{4} = -3 \) =======> Correct

Critical Numbers: \(\displaystyle \dfrac{3}{2}, -3\) =======> Correct

Find Max Min

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 = -\dfrac{370}{8}\) =======> Incorrect (arithmetic)

\(\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 215 \) =======> Incorrect (arithmetic)

Where does the interval increase or decrease?

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

For \(\displaystyle x < -3\)

\(\displaystyle f'(-4) = 12(-4)^{2} + 18(-4) - 54 = \)

\(\displaystyle f'(-4) = 192 -72 - 54 = 66 \) Positive so increasing =======> Correct

For \(\displaystyle -3 < x <\dfrac{3}{2}\)

\(\displaystyle f'(-1) = 12(-1)^{2} + 18(-1) - 54 = \)

\(\displaystyle f'(-1) = 12 - 18 - 54 = -60\) Negative so decreasing =======> Correct

For \(\displaystyle \dfrac{3}{4} < x\) =======> Incorrect (should be x > 3/2)
\(\displaystyle f'(1) = 12(1)^{2} + 18(1) - 54 = - 24 \) Negative so decreasing =======> Incorrect (1 can't be a test point since it is less than 3/2)

What is the infection point?

\(\displaystyle f''(x) = 24x + 18\)

\(\displaystyle 24x + 18 = 0\) =======> Correct

\(\displaystyle 24x = -18\) =======> Correct

\(\displaystyle x = -\dfrac{4}{3}\) =======> Incorrect (arithmetic)

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{4}{3})^{3} + 9(-\dfrac{4}{3})^{2} - 54(-\dfrac{4}{3}) + 8 \) =======> Incorrect (because your POI is wrong)

\(\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{64}{67}) + 9(\dfrac{16}{9}) - 54(-\dfrac{4}{3}) + 8 \) =======> Incorrect (because your POI is wrong)

\(\displaystyle f(-\dfrac{4}{3}) = -\dfrac{256}{67}) + 16 + \dfrac{216}{3}) + 8 \) :confused: =======> Incorrect (because your POI is wrong)
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\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle f''(x) = 24x + 18\)

Find Critical Numbers:

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x + 3x - 9) = 0\)

\(\displaystyle \dfrac{-3 \pm \sqrt{3^{2} - 4(2)(-9)}}{2(2)}\)

\(\displaystyle \dfrac{-3 \pm \sqrt{81}}{4}\)

\(\displaystyle \dfrac{-3 \pm 9}{4}\)

\(\displaystyle \dfrac{-3 + 9}{4} = \dfrac{6}{4} = \dfrac{3}{2} \)

\(\displaystyle \dfrac{-3 - 9}{4} = \dfrac{-12}{4} = -3 \)

Critical Numbers: \(\displaystyle \dfrac{3}{2}, -3\)

Find Max Min

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 =43\) ;)

\(\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 287 \) ;)

Where does the interval increase or decrease?

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

For \(\displaystyle x < -3\)

\(\displaystyle f'(-4) = 12(-4)^{2} + 18(-4) - 54 = \)

\(\displaystyle f'(-4) = 192 -72 - 54 = 66 \) Positive so increasing

For \(\displaystyle -3 < x <\dfrac{3}{2}\)

\(\displaystyle f'(-1) = 12(-1)^{2} + 18(-1) - 54 = \)

\(\displaystyle f'(-1) = 12 - 18 - 54 = -60\) Negative so decreasing

For \(\displaystyle x >\dfrac{3}{2}\)

\(\displaystyle f'(1) = 12(3)^{2} + 18(3) - 54 = 108 \) Positive so increasing ;)

What is the infection point?

\(\displaystyle f''(x) = 24x + 18\)

\(\displaystyle 24x + 18 = 0\)

\(\displaystyle 24x = -18\)

\(\displaystyle x = -\dfrac{4}{3}\) :?: This looks right to me cause \(\displaystyle 6\) goes into \(\displaystyle 24\), \(\displaystyle 6\) times and \(\displaystyle 6\) goes into \(\displaystyle 18\), \(\displaystyle 3\) times.

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

\(\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{4}{3})^{3} + 9(-\dfrac{4}{3})^{2} - 54(-\dfrac{4}{3}) + 8 \)

\(\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{64}{67}) + 9(\dfrac{16}{9}) - 54(-\dfrac{4}{3}) + 8 \)

\(\displaystyle f(-\dfrac{4}{3}) = -\dfrac{256}{67}) + 16 + \dfrac{216}{3}) + 8 \) :confused:
 
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In regards to these:


Jason76 said:
\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 \)

Find Critical Numbers:

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x + 3x - 9) = 0 \ \ \ \ \ \)Wrong. There is a typo. You are missing an exponent of "2" on the first x term.


\(\displaystyle 2x^2 + 3x - 9 = 0 \ \ \ \ \) This is the quadratic equation in question.


x = \(\displaystyle \ \dfrac{-(3) \pm \sqrt{(3)^{2} - 4(2)(-9)}}{2(2)} \ \ \ \ \ \)Put grouping symbols consistently around everything being

substituted for.


x = \(\displaystyle \ \dfrac{-3 \pm \sqrt{9 + 72}}{4} \ \ \ \ \ \) optional intermediate step


x = \(\displaystyle \ \dfrac{-3 \pm \sqrt{81}}{4}\)


x = \(\displaystyle \ \dfrac{-3 \pm 9}{4}\)


x = \(\displaystyle \ \dfrac{-3 + 9}{4} = \dfrac{6}{4} = \dfrac{3}{2} \)


x = \(\displaystyle \ \dfrac{-3 - 9}{4} = \dfrac{-12}{4} = -3 \)



Critical Numbers: \(\displaystyle \ \ \) x = \(\displaystyle \ \dfrac{3}{2}, -3\)
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Again, you must include the "x =" or that portion is wrong.
 
Found maximum to be \(\displaystyle 143\) (from the \(\displaystyle -3\) critical point), and it's correct on computer. But finding the minimum with the \(\displaystyle \dfrac{3}{2}\) critical point is tough. Any hint? :confused:
 
Jason76 said:
Found maximum to be \(\displaystyle 143\) (from the \(\displaystyle -3\) critical point), and it's correct on computer. But finding the minimum with the \(\displaystyle \dfrac{3}{2}\) critical point is tough. Any hint? :confused:
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Jason, it is arithmetic! You know how to do Calculus (for the most part), but you can't figure out what f(x) = 4x^3 + 9x² - 54x +8 is at x = 3/2?

The hint is to plug in x = 3/2 where you see x, cube it, square it, etc then get a common denominator and wallah, you have your value. Try it and get back to us with your answer.
 
Inflection Point

\(\displaystyle f''(x) = 36x^{2} + 18 \)

\(\displaystyle 36x^{2} + 18 = 0\)

\(\displaystyle 18(2x^{2} + 1) = 0\)

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

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

\(\displaystyle x^{2} = -\dfrac{1}{2}\) :confused:
 
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Jason76 said:
Inflection Point

\(\displaystyle f''(x) = 36x^{2} + 18 \)

\(\displaystyle 36x^{2} + 18 = 0\)

\(\displaystyle 18(2x^{2} + 1) = 0\)

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

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

\(\displaystyle x^{2} = -\dfrac{1}{2}\) :confused:
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Why did you even post that, Jason76!? It's not part of this problem.
 
And, in any case, \(\displaystyle 36x^2+ 18\), for all x, is a sum of positive numbers so is NEVER 0.
 
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