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Vlad1618 [11]
1 year ago
5

I did A and B I just need help with the rest!

Mathematics
1 answer:
Rom4ik [11]1 year ago
6 0

Solution

(a) setting f(x) = 0

\begin{gathered} 3x^4-18x^3-21x^2+144x-108=0 \\  \\ (x+3)(x-1)(x-2)(x-6)=0 \\  \\ \Rightarrow x=-3,1,2,6 \end{gathered}

(b)

when x = 0

y = f(0) = -108

Hence, the height is 108

(c)

Maximum = (1.5, 15.188)

(d)

Minimum (-1.702, -300) and (4.702, -300)

(e) From the graph, the interval of increasing is

\begin{gathered} (-1.7,\frac{3}{2}),(4.7,\infty) \\  \end{gathered}

(f) From the graph, the interval of decreasing is;

(-\infty,-1.7),(\frac{3}{2},4.7)

g)

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Let v⃗ 1=⎡⎣⎢033⎤⎦⎥,v⃗ 2=⎡⎣⎢1−10⎤⎦⎥,v⃗ 3=⎡⎣⎢30−3⎤⎦⎥ be eigenvectors of the matrix A which correspond to the eigenvalues λ1=−1, λ2
kaheart [24]

Answer:

- x as a linear combination :

x = -1 v1+ 0 v2+ 1 v3.

- Transpose Ax = (12, -6, -6)

Step-by-step explanation:

Given v1 = (0 3 3),v2 = (1 −1 0), v3 = (3 0 −3) be eigenvectors of the matrix A which correspond to the eigenvalues λ1 = −1, λ2 = 0, and λ3 = 1, respectively, and let x = (−2 −4 0). Express x as a linear combination of v1, v2, and v3, and find Ax .

To write x as a linear combination of v1, v2, and v3

x = -1 v1+ 0 v2+ 1 v3.

To find Ax

Write A = (0 ......3 ......3 )

...................(1 ......-1 ......0)

...................(3 ......0......-3)

Since transpose x = (-2, 4, 0)

Ax =......... (0 ......3......3 )(-2)

...................(1 ......-1 ......0)(4)

...................(3 ......0......-3)(0)

= (0×-2 + 3×4 + 3×0)

...(1×-2 + -1×4 + 0×0)

.. (3×-2 + 0×4 + -3×0)

As = (12)

....(-6)

....(-6)

Transpose Ax = (12, -6, -6)

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3 years ago
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oee [108]

Answer:

a = ln 35

Step-by-step explanation:

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