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Crank
3 years ago
5

Can anyone help me find the inflection point and how to solve this cubic function?

Mathematics
1 answer:
Anna71 [15]3 years ago
5 0

Answer:

x-intercept → (-4, 0)

y-intercept → (0, -2.44)

Domain: (-∞, ∞)

Range: (-∞, ∞)

Step-by-step explanation:

Given function is f(x) = -\sqrt[3]{(x+3)}-1

For x-intercept,

-\sqrt[3]{(x+3)}-1=0

-\sqrt[3]{(x+3)}=1

\sqrt[3]{(x+3)}=-1

(x + 3) = -1

x = -4

Therefore, x-intercept → (-4, 0)

For y-intercept,

Substitute x = 0 in the function.

f(x) = -\sqrt[3]{(0+3)}-1

     = -\sqrt[3]{3}-1

     = -1.44 - 1

     = -2.44

Therefore, y-intercept → (0, -2.44)

Domain: This function is defined for all real values of x.

Therefore, Domain: (-∞, ∞)

Range: Since this function is defined for all real values of x, we will get a distinct output value for every distinct input values.

Therefore, Range: (-∞, ∞)

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If a = pi +3j - 7k, b = pi - pj +4k and the angle between a and is acute then the possible values for p are given by​
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The family of possible values for p are:

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Step-by-step explanation:

By Linear Algebra, we can calculate the angle by definition of dot product:

\cos \theta = \frac{\vec a\,\bullet\,\vec b}{\|\vec a\|\cdot \|\vec b\|} (1)

Where:

\theta - Angle between vectors, in sexagesimal degrees.

\|\vec a\|, \|\vec b \| - Norms of vectors \vec {a} and \vec{b}

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Maximum (\cos \theta = 1)

\frac{p^{2}-3\cdot p -28}{\sqrt{p^{2}+58}\cdot \sqrt{2\cdot p^{2}+16}} < 1

With the help of a graphing tool we get the family of possible values for p are:

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2 years ago
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