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Harman [31]
3 years ago
15

At what value(s) of x does f(x) = –x^4+2x^2 have a critical point where the graph changes from decreasing to increasing?

Mathematics
1 answer:
Delicious77 [7]3 years ago
7 0

Answer:

f''(-1) and f''(1) are both negative, so at x=-1 and x=1, the function is changing from increasing to decreasing.

Step-by-step explanation: We usually indicate exponents with the "^" symbol. This makes the function f(x) = -x^4 + 2x^2.

f'(x) = -4x^3 + 4x; f'(x) = 0 when x= 0 or ±1.

f''(x) = -12x^2 + 4, and this is > 0 when x=0, so x=0 is a relative minimum point on the graph. This means that the function is changing from decreasing to increasing.

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2 years ago
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Answer:

Solution given:

<u>A.coordinate are</u>

A(-2,3)

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C(4,5)

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we have

length \sqrt{(x2-x1)²+(y2-y1)²}

now

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AC:\sqrt{(-2-4)²+(3-5)²}=2\sqrt{10}units

<u>C.</u><u> the </u><u>figure</u><u>:</u>

<u>By</u><u> </u><u>using</u><u> </u><u>Pythagoras</u><u> </u><u>law</u>

base[b]=AB=perpendicular [p]=AC

hypotenuse [h]=BC

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h²=p²+b²

substituting value

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80=2*4*10

80=80

<u>SO</u><u> </u><u>IT</u><u> </u><u>IS</u><u> </u><u>RIGHT</u><u> </u><u>ANGLED</u><u> </u><u>ISOSCELES</u><u> </u><u>TRIANGLE</u><u>.</u>

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Answer:

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Step-by-step explanation:

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7 0
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