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3 years ago

Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = ln(x2 + 3x + 4), [−2, 2]

Mathematics

1 answer:

vivado [14]3 years ago

4 0

The vertex (minimum) of the quadratic ax² +bx +c is located at x=-b/(2a). This means the minimum value of f(x) will be found at x = -3/(2*1) = -1.5.

Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.

On the given interval, ...
the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057

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Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, sphere

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3 years ago

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Answer:

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Add the 2 opposite angles to find the exterior angle.

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Mkey [24]

Answer:

Table C

Step-by-step explanation:

we know that

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Table A

For $x=-2, y=-4$ ------> $k=-4/-2=2$

For $x=-1, y=-3$ ------> $k=-3/-1=3$

This table has different values of k

therefore

the table A does not represent a proportional relationship

Table B

For $x=-2, y=-4$ ------> $k=-4/-2=2$

For $x=-1, y=-2$ ------> $k=-2/-1=2$

For $x=1, y=3$ ------> $k=3/1=3$

This table has different values of k

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the table B does not represent a proportional relationship

Table C

For $x=-2, y=-4$ ------> $k=-4/-2=2$

For $x=-1, y=-2$ ------> $k=-2/-1=2$

For $x=1, y=2$ ------> $k=2/1=2$

For $x=2, y=4$ ------> $k=4/2=2$

This table has the same value of k

therefore

the table C represent a proportional relationship

Table D

For $x=-2, y=-4$ ------> $k=-4/-2=2$

For $x=-1, y=-2$ ------> $k=-2/-1=2$

For $x=1, y=2$ ------> $k=2/1=2$

For $x=2, y=6$ ------> $k=6/2=3$

This table has different values of k

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Answer:

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Step-by-step explanation:

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