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Tatiana [17]
2 years ago
6

Find f( 2 ) , f( 3 ) , f( 4 ) , and f( 5 ) if f is defined recur-sively by f( 0 ) = f( 1 ) = 1 and for n = 1 , 2 ,... a) f(n + 1

) = f(n) − f(n − 1 ) . b) f(n + 1 ) = f(n)f(n − 1 ) . c) f(n + 1 ) = f(n) 2 + f(n − 1 ) 3 . d) f(n + 1 ) = f(n)/f(n − 1 ) . slader
Mathematics
1 answer:
irakobra [83]2 years ago
7 0

Answer:

(a) f(2) = 0, f(3) = -1, f(4) = -1, f(5) = 0

(b) f(2) = 1, f(3) = 1, f(4) = 1, f(5) = 1

(c) f(2) = 5, f(3) = 13, f(4) = 41, f(5) = 121

(d) f(2) = 1, f(3) = 1, f(4) = 1, f(5) = 1

Step-by-step explanation:

(a)

f(2) = f(1) - f(0) = 1 - 1 = 0

f(3) = f(2) - f(1) = 0 - 1 = -1

f(4) = f(3) - f(2) = -1 - 0 = -1

f(5) = f(4) - f(3) = -1 - (-1) = 0

(b)

f(2) = f(1)*f(0) = 1*1 = 1

f(3) = f(2)*f(1) = 1*1 = 1

f(4) = f(3)*f(2) = 1*1 = 1

f(5) = f(4)*f(3) = 1*1 = 1

(c)

f(2) = 2f(1) + 3f(0) = 2 + 3 = 5

f(3) = 2f(2) + 3f(1) = 10 + 3 = 13

f(4) = 2f(3) + 3f(2) = 26 + 15 = 41

f(5) = 2f(4) + 3f(3) = 82 + 39 = 121

(d)

f(2) = f(1)/f(0) = 1/1 = 1

f(3) = f(2)/f(1) = 1/1 = 1

f(4) = f(3)/f(2) = 1/1 = 1

f(5) = f(4)/f(3) = 1/1 = 1

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Step-by-step explanation:

<u>Optimizing With Derivatives </u>

The procedure to optimize a function (find its maximum or minimum) consists in :

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We know a cylinder has a volume of 4 ft^3. The volume of a cylinder is given by

\displaystyle V=\pi r^2h

Equating it to 4

\displaystyle \pi r^2h=4

Let's solve for h

\displaystyle h=\frac{4}{\pi r^2}

A cylinder with an open-top has only one circle as the shape of the lid and has a lateral area computed as a rectangle of height h and base equal to the length of a circle. Thus, the total area of the material to make the cylinder is

\displaystyle A=\pi r^2+2\pi rh

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\displaystyle A=\pi r^2+2\pi r \left (\frac{4}{\pi r^2}\right )

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\displaystyle 2\pi r=\frac{8}{r^2}

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\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet

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\displaystyle h=\frac{4}{\pi \ r^2}\approx 1.084\ feet

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\displaystyle A''=2\pi+\frac{16}{r^3}

We can see it will be always positive regardless of the value of r (assumed positive too), so the critical point is a minimum.

The minimum area is

\displaystyle A=\pi(1.084)^2+\frac{8}{1.084}

\boxed{ A=11.07\ ft^2}

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