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bezimeni [28]
2 years ago
7

The function f(x)=2/3x-4 was replaced with f(x+k)=2/3x-6 what is the value of k

Mathematics
1 answer:
HACTEHA [7]2 years ago
7 0

Answer:

The given function f(x)=\frac{2x}{3}-4 was replaced with f(x+k)=\frac{2x}{3}-6 then the value of k is -3

that is k=-3.

Step-by-step explanation:

The given function is f(x)=\frac{2x}{3}-4\hfill (1)

and when x=x+k then f(x+k)=\frac{2x}{3}-6\hfill (2)

Now to find the value of k:

Put x=x+k in the function f(x)

f(x)=\frac{2x}{3}-4

f(x+k)=\frac{2(x+k)}{3}-4

f(x+k)=\frac{2x+2k}{3}-4\hfill (3)

Now comparing equations (2) and (3) we get

f(x+k)=\frac{2x+2k}{3}-4=\frac{2x}{3}-6

\frac{2x+2k}{3}-4=\frac{2x}{3}-6

\frac{2x+2k}{3}-4-\frac{2x}{3}+6=0

\frac{2x}{3}+\frac{2k}{3}-4-\frac{2x}{3}+6=0

\frac{2k}{3}+2=0

\frac{2k}{3}=-2

k=\frac{-2\times 3}{2}

k=-3

Therefore k=-3

Therefore the given function f(x)=\frac{2x}{3}-4 was replaced with  f(x+k)=\frac{2x}{3}-6 then the value of k is -3

that is k=-3.

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∫ 17 / (x³ − 125) dx

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Use partial fraction decomposition:

= 17 ∫ [ A / (x − 5) + (Bx + C) / (x² + 5x + 25) ] dx

Use common denominator to find the missing coefficients.

A (x² + 5x + 25) + (Bx + C) (x − 5) = 1

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Match the coefficients and solve the system of equations.

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5A − 5B + C = 0

25A − 5C = 1

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B = -1/75

C = -2/15

So the integral is:

= 17 ∫ [ 1/75 / (x − 5) + (-1/75 x − 2/15) / (x² + 5x + 25) ] dx

Simplify:

= 17/75 ∫ [ 1 / (x − 5) − (x + 10) / (x² + 5x + 25) ] dx

Factor ½ from the numerator of the second fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 20) / (x² + 5x + 25) ] dx

Split the fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − ½ (15) / (x² + 5x + 25) ] dx

Multiply the last fraction by 4/4:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 30 / (4x² + 20x + 100) ] dx

Complete the square:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 15 / ((2x + 5)² + 75) ] dx

Split the integral:

= 17/75 ∫ 1 / (x − 5) dx − 17/150 ∫ (2x + 5) / (x² + 5x + 25) dx − 17/5 ∫ 1 / ((2x + 5)² + 75) dx

The first integral is:

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The second integral is:

∫ (2x + 5) / (x² + 5x + 25) dx = ln(x² + 5x + 25)

The third integral is:

∫ 1 / ((2x + 5)² + 75) dx = 1/√75 tan⁻¹((2x + 5) / √75)

Plug in:

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