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2 years ago

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F(x) = x². What is g(x)?

1 answer:

igor_vitrenko [27]2 years ago

4 0

The function f(x) = x² then the required function exists g(x) = -x²- 4.

<h3>What is a function?</h3>

The function exists described as y = f(x).

In mathematics, a function from a set X to a set Y allocates to each element of X exactly one element of Y. The set X exists named the domain of the function and the set Y exists named the codomain of the function. Functions stood originally for the idealization of how a variable quantity relies on another quantity.

For every x there exists a certain value of y.

From the graph g(x) exists reflection of f(x) at y = -4.

So g(x) = -f(x) - 4, the negative sign for reflection.

g(x) = -x² - 4

The required function exists g(x) = -x² - 4.

Therefore, the correct answer is option D. g(x) = -x² - 4.

To learn more about functions, refer to:

brainly.com/question/12431044

#SPJ9

The complete question is:

F(x) = x². What is g(x)?

A. g(x) = x² - 4

B. g(x) = x² + 4

C. g(x) = -4x²

D. g(x) = -x² - 4

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C=1/21.22.23+1/22.23.24+................+1/200.201.202<br><br> . = là dấu nhân

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It looks like you have to find the value of the sum,

$C = \displaystyle \frac1{21\times22\times23} + \frac1{22\times23\times24} + \cdots + \frac1{200\times201\times202}$

so that the <em>n</em>-th term in the sum is

$\dfrac1{(21+(n-1))\times(21+n)\times(21+(n+1))} = \dfrac1{(n+20)(n+21)(n+22)}$

for 1 ≤ <em>n</em> ≤ 180.

We can then write the sum as

$\displaystyle C = \sum_{n=1}^{180} \frac1{(n+20)(n+21)(n+22)}$

Break up the summand into partial fractions:

$\dfrac1{(n+20)(n+21)(n+22)} = \dfrac a{n+20} + \dfrac b{n+21} + \dfrac c{n+22}$

Combine the fractions into one with a common denominator and set the numerators equal to one another:

$1 = a(n+21)(n+22) + b(n+20)(n+22) + c(n+20)(n+21)$

Expand the right side and collect terms with the same power of <em>n</em> :

$1 = a(n^2+43n+462)+b(n^2+42n+440) + c(n^2+41n + 420) \\\\ 1 = (a+b+c)n^2 + (43a+42b+41c)n + 462a+440b+420c$

Then

<em>a</em> + <em>b</em> + <em>c</em> = 0

43<em>a</em> + 42<em>b</em> + 41<em>c</em> = 0

462<em>a</em> + 440<em>b</em> + 420<em>c</em> = 1

==> <em>a</em> = 1/2, <em>b</em> = -1, <em>c</em> = 1/2

Now our sum is

$\displaystyle C = \sum_{n=1}^{180} \left(\frac1{2(n+20)}-\frac1{n+21}+\frac1{2(n+22)}\right)$

which is a telescoping sum. If we write out the first and last few terms, we have

<em>C</em> = 1/(2×21) - 1/22 <u>+ 1/(2×23)</u>

… … + 1/(2×22) - 1/23 <u>+ 1/(2×24)</u>

… … <u>+ 1/(2×23)</u> - 1/24 <u>+ 1/(2×25)</u>

… … <u>+ 1/(2×24)</u> - 1/25 <u>+ 1/(2×26)</u>

… … + … - … + …

… … <u>+ 1/(2×198)</u> - 1/199 <u>+ 1/(2×200)</u>

… … <u>+ 1/(2×199)</u> - 1/200 + 1/(2×201)

… … <u>+ 1/(2×200)</u> - 1/201 + 1/(2×202)

Notice the diagonal pattern of underlined and bolded terms that add up to zero (e.g. 1/(2×23) - 1/23 + 1/(2×23) = 1/23 - 1/23 = 0). So, like a telescope, the sum collapses down to a simple sum of just six terms,

<em>C</em> = 1/(2×21) - 1/22 + 1/(2×22) + 1/(2×201) - 1/201 + 1/(2×202)

which we simplify further to

<em>C</em> = 1/42 - 1/44 - 1/402 + 1/404

<em>C</em> = 1,115/1,042,118 ≈ 0.00106994

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