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Gala2k [10]
3 years ago
15

Problem 1.9. Find the relative ertrema of the following function. Determine whether each

Mathematics
1 answer:
Igoryamba3 years ago
6 0

Answer:

Min: 0

Max: 0.125

Step-by-step explanation:

\frac{d}{dx} \frac{x^2}{x^4+16} =0\\\frac{-(2x)(x^4-16)}{(x^4+16)^2} =0\\x=0, 2, -2\\f(0)=0\\f(2)=f(-2)=0.125\\\frac{d^2}{dx^2} f(x)=\frac{6x^8-384x^4+512}{(16+x^4)^3} \\f''(0)=0.125>0\\f''(2)=f''(-2)=-0.125

Min: 0

Max: 0.125

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Graph triangle RST with vertices R(3, 7), S(-5, -2), and T(3, -5) and its image after a reflection over x = -3.​
kirill115 [55]

Given:

The vertices of a triangle are R(3, 7), S(-5, -2), and T(3, -5).

To find:

The vertices of the triangle after a reflection over x = -3 and plot the triangle and its image on the graph.

Solution:

If a figure reflected across the line x=a, then

(x,y)\to (-(x-a)+a,y)

(x,y)\to (-x+a+a,y)

(x,y)\to (2a-x,y)

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(x,y)\to (2(-3)-x,y)

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S(-5,-2)\to S'(-6+5,-2)

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And,

T(3,-5)\to T'(-6-3,-5)

T(3,-5)\to T'(-9,-5)

Therefore, the vertices of triangle after reflection over x=-3 are R'(-9,7), S'(-1,-2) and T'(-3,-5).

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Answer:

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Step-by-step explanation:

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