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Evgesh-ka [11]
3 years ago
7

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

fy your answer using an analysis of f ′(x) and f ′′(x)
Mathematics
1 answer:
stealth61 [152]3 years ago
8 0
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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sasho [114]
Answer is 54^2.

since there are triangles in this one, it is easier to do this. all you have to do is make boxes in and since there are angles, outside the lines. like this picture.

the orange is 1
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blue is 3
purple is 4(a regular box which should be easy to count.

all you have to do is add up all the boxes within the boundaries.
boundary 1: 12
boundary 2: 18
boundary 3: 6
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now when the boundary has a triangle in it (1, 2, & 3) divide the number you got in half or 2.
boundary 1: 6
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the ractangle box doesn't not get divided.
boundary 1: 6
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add all the numbers you got now for each boundary and that would be your area squared.

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Step-by-step explanation:

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