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ANEK [815]
3 years ago
10

Evaluate f (x) = 2x4 – 4x3 – 11x2 3x – 6 for x = –2. f (–2) =

Mathematics
2 answers:
Musya8 [376]3 years ago
8 0
Hello,

After decrypting the question:

f(x)=2x^4-4x^3-11x^2+3x-6

f(-2)=2(-2)^4-4(-2)^3-11(-2)^2+3*(-2)-6=2*16+32-44-6-6=8

Art [367]3 years ago
5 0

Answer:

The value of f(x) is 8 at x=-2.

Step-by-step explanation:

The given function is

f(x)=2x^4-4x^3-11x^2+3x-6

We have to find the value of function at x=-2.

Substitute x=-2 in the given function.

f(-2)=2(-2)^4-4(-2)^3-11(-2)^2+3(-2)-6

f(-2)=32+32-44-6-6

f(-2)=64-56

f(-2)=8

Therefore the value of f(x) is 8 at x=-2.

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How to solve this trigonometric equation cos3x + sin5x = 0
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Answer:

  x = {nπ -π/4, (4nπ -π)/16}

Step-by-step explanation:

It can be helpful to make use of the identities for angle sums and differences to rewrite the sum:

  cos(3x) +sin(5x) = cos(4x -x) +sin(4x +x)

  = cos(4x)cos(x) +sin(4x)sin(x) +sin(4x)cos(x) +cos(4x)sin(x)

  = sin(x)(sin(4x) +cos(4x)) +cos(x)(sin(4x) +cos(4x))

  = (sin(x) +cos(x))·(sin(4x) +cos(4x))

Each of the sums in this product is of the same form, so each can be simplified using the identity ...

  sin(x) +cos(x) = √2·sin(x +π/4)

Then the given equation can be rewritten as ...

  cos(3x) +sin(5x) = 0

  2·sin(x +π/4)·sin(4x +π/4) = 0

Of course sin(x) = 0 for x = n·π, so these factors are zero when ...

  sin(x +π/4) = 0   ⇒   x = nπ -π/4

  sin(4x +π/4) = 0   ⇒   x = (nπ -π/4)/4 = (4nπ -π)/16

The solutions are ...

  x ∈ {(n-1)π/4, (4n-1)π/16} . . . . . for any integer n

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