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Archy [21]
3 years ago
5

Do I have this one correct

Mathematics
1 answer:
nikklg [1K]3 years ago
8 0

Answer:

Option B is correct

Step-by-step explanation:

Given is a cos funciton.

Period = 3

Max = 20 and minimum =0

Hence centre line is y = 1/2 (max+min)

i.e. y =10 is the central line

Let the funciton be of the form

y = a cos (bx+C)+d

We got d = 10...i

Period =3

Hence b = \frac{2\pi}{3} ... ii

Amplitude = a = distance of max from central line = 20-10 =10...iii

Since reflection over x axis, sign of a =negative

a=-10

Since no horizontal shift, c=0 ... iv

Put to gether the function is

y=-10cos\frac{2\pi x}{3} +10

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Answer:

As consequence of the Taylor theorem with integral remainder we have that

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt

If we ask that f has continuous (n+1)th derivative we can apply the mean value theorem for integrals. Then, there exists c between a and x such that

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x

Hence,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .

Thus,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

and the Taylor theorem with Lagrange remainder is

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.

Step-by-step explanation:

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