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Andrew [12]
3 years ago
8

Given the function

alt="g(x)=m(-2x+2)^5-n" align="absmiddle" class="latex-formula"> where m ≠ 0 and n ≠ 0 are constants.
A. Prove that g is monotonic (this means that either g always increases or g always decreases)

B. Show that the x-coordinate(s) of the location(s) of any critical points are independent of m and n
Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
4 0

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

The given function g(x) is a continuous function, since, for any x we can find a real value of the function.

A. \frac{d g(x)}{dx} = 5m(-2x + 2)^{4} \times (-2).

Since, m is a constant, which is not equals to 0, the above value of the differentiation of the function, will be negative.

For x = 1, the above value is 0, that is at x =  1, the function has either maximum value, or a minimum value.

B. As per the above information, we have get that for x = 1, \frac{d g(x)}{dx} = 0.

Hence, the function's critical point's x coordinate is x = 1.

The x- coordinate of the given point is not dependent on m or n.

Hence, proved.

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(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)}  \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)}  -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\

\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}

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In a triangle, the centroid divides the median in the ratio 2:1.

It is given that AD is the median and AD = 33

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Therefore, P divides AD in the ratio 2:1.

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