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max2010maxim [7]
3 years ago
14

What is the end behavior of a sixth degree polynomial function whose leading coefficient is

Mathematics
1 answer:
bija089 [108]3 years ago
5 0

Answer:

  both ends tend toward negative infinity

Step-by-step explanation:

The ends of an even-degree polynomial go in the same direction. The sign of that direction matches the sign of the leading coefficient.

Here, the leading coefficient is negative, so ...

  y → -∞   when   x → ±∞

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Answer:

b AND e

Step-by-step explanation:

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3 years ago
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Jimmy has listed the amount of money in his wallet for each of the last ten days. He decides to remove day 7, as that was payday
abruzzese [7]

Answer:

The mean will reduce

Step-by-step explanation:

The total amount of money for the ten days divided by number of days will give mean 1, M₁

Removing the value for day seven will reduce the sum of money as it will be for 9 days only.In finding the mean, you will divide by 9 days to get mean M₂

Mean M₂ < M₁

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Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi
stealth61 [152]
Applying our power rule gets us our first derivative,

\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}

simplifying a little bit,

\rm f'(x)=2x^{-2/3}+4x^{1/3}

looking for critical points,

\rm 0=2x^{-2/3}+4x^{1/3}

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

0=2x^{-2/3}\left(1+2x\right)

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

\rm f'(x)=2x^{-2/3}+4x^{1/3}

and take our second derivative.

\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}

Looking for inflection points,

\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}

Again, pulling out the smaller power of x, and fractional part,

\rm 0=-\frac43x^{-5/3}\left(1-x\right)

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
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