1)If f is a function such that f(1)=2 and f(n+1)=3f(n)+1/3 for n=1,2,3..., what is the value of f(100)?

Question

Please show your steps! Really need help! ><

Thank you so much!

2)Determine the quadratic function f(x)=ax^2+bx+c, such that f(x)+2x+8 = f(x+1), and f(1)=12.

3) For what values of x is the expression (|x|-1)(|x|+2) negative.

Thank You!

Answer 1

1) You can look for such f(n) that f(n) = a*3^n + b and find from f(n+1) = a*3^(n+1) + b = 3*f(n) + 1/3 = a*3^(n+1) + 3b + 1/3 that b = 3b + 1/3, i.e. b = -1/6. So f(n) = a*3^n - 1/6. You can find a from f(1) = 3a - 1/6 = 2. a = 13/18. Finally f(n) = 13/18*3^n - 1/6
f(100) = 13/18*3^100 - 1/6

2) ax^2 + bx + c + 2x + 8 = a(x + 1)^2 + b(x + 1) + c
2x + 8 = 2ax + a + b
2a = 2
a + b = 8
a = 1, b = 7
f(1) = (a + b) + c = 8 + c = 12
c = 4

3) |x| + 2 > 0. So we need to find |x| - 1 < 0, or |x| < 1, or - 1 < x < 1

Answer 2

1) Given f(1)=2 and f(n+1)=3f(n)+1/3 for n=1,2,3...,

So
f(2) = 3f(1) + 1/3,

f(3) = 3f(2) + 1/3 = 3[3f(1) + 1/3] + 1/3
f(3) = (3^2) f(1) + 1+ 1/3

f(4) = 3f(3) + 1/3
f(4) = 3[ (3^2) f(1) + 1+ 1/3] + 1/3
f(4) = (3^3) f(1) + 3 + 1 + 1/3

f(5) = 3f(4) + 1/3
f(5) = 3[(3^3) f(1) + 3 + 1 + 1/3] + 1/3
f(5) = (3^4) f(1) + (3^2) + 3 + 1 + 1/3

f(n+1) = f(n) + 1/3
f(n+1) = (3^n)f(1) + [3^(n-2) + 3^(n-3) + ...+ 1 + 1/3]

where [3^(n-2) + 3^(n-3) + ...+ 1 + 1/3] is geometric series with (n -1)th term (inclusive 1 and 1/3) a = 1/3, r = 3
Thus

S(n-1) = (1/3) [ 3^n - 1] / (3 - 1)
S(n-1) = (1/6) [ 3^n - 1]

Finally,
f(n+1) = (3^n)f(1) + (1/6) [ 3^n - 1]

So
f(100) = [3^99] f(1) + (1/6) [3^99 - 1]
f(100) = [3^99] (2) + (1/6) [3^99 - 1]
f(100) = [3^99] (2+1/6) - 1/6
f(100) = [3^98] (13/2) - 1/6 = [3^98] (13/2)

as 1/6 is too small to deduct from [3^98] (13/2)

2) f(x) = ax^2 + bx + c

Given f(1)=12, then a+b+c = 12 ...(1)

f(x+1) = a(x+1)^2 + b(x+1) + c
f(x+1) = ax^2+2ax+a +bx+ b +c

Given f(x)+2x+8 = f(x+1)
ax^2 + bx + c + 2x + 8 = ax^2+2ax+a +bx+ b +c
(b+2)x + (c+8) = (2a+b)x + (a+b+c) ...(2)

Replace (1) into (2)
(b+2)x + (c+8) = (2a+b)x + 12

So 2a = 2, a =1
c + 8 = 12, c = 4
and b = 7.

Thus f(x) = x^2+7x+4

3) From expression (|x|-1)(|x|+2),
(|x|+2) is always positive.
However (|x|-1) is negative, if and only if
|x|-1 < 0
|x| < 1
or equal to
-1< x < 1

Answer 3

1) I disagree with Dr. D's conception that you've typed in the recurrence relation incorrectly. It seems unlikely to me that a recurrence relation would be given in his form, and the resulting relation seems a bit too simple.

As given, we have
f(n+1) = 3f(n) + 1/3
f(n+2) = 9f(n) + 1 + 1/3
...
f(n+k) = 3^k f(n) + 3^(k-1) + 3^(k-2) + ... + 1 + 1/3.
So f(100) = f(1 + 99)
= 3^99 f(1) + 3^98 + 3^97 + ... + 1 + 1/3.
Now we know that 1/3 + 1 + ... + 3^98
= 1/3 (3^100 - 1) / (3 - 1)
(sum of a geometric series)
= 3^99 / 2 - 1/6.

So f(100) = 3^99 (2) + 3^99 / 2 - 1/6
= (3/2) 3^99 - 1/6
= 3^100 / 2 - 1/6; or, if you prefer
= (3^100 - 1/3) / 2.

2) f(x) = ax^2 + bx + c
f(x+1) = f(x) + 2x + 8
<=> a(x+1)^2 + b(x+1) + c = ax^2 + bx + c + (2x + 8)
<=> ax^2 + (b+2a)x + (a+b+c) = ax^2 + (b+2)x + (c+8)

So we get b+2a = b+2 <=> a = 1, and a + b + c = c + 8.
f(1) = 12 => a + b + c = 12
So c + 8 = 12 and hence c = 4. So b = 12 - 1 - 4 = 7.
So f(x) = x^2 + 7x + 4.

3). |x| + 2 is always strictly positive. So the expression is negative precisely when |x| - 1 < 0, i.e. |x| < 1, i.e. -1 < x < 1.

Answer 4

Quad and ksoileau have already incredibly a lot crushed the main out of this question. So permit me merely make a pair of feedback: a) in case you require continuity, then there is precisely one function that fulfill your situation. this is because of the fact i) Your situation implies right now that f is invertible (f^{-a million}(x) = f(f(x))) so ii) f is a homeomorphism of the actual line and f preserves order (because of the fact the actual line is an ordered set.) so iii) if f(a) = b for b no longer equivalent to a for some a, then we are able to, without loss of generality, anticipate that f(a) < a. Then f(f(a)) < f(a) < a, and in addition f(f(f(a))) < a. b) As quad has confirmed, it suffices to have one discontinuity. c) we are able to take additionally ksoileau's attitude slightly greater. Given a function f, we are able to evaluate the curve C in R^3 given via (x,f(x),f(f(x))). Requiring that f(f(f(x))) = x is then the comparable as requiring that C is invariant below the action of the rotation matrix [ 0 0 a million ] [ a million 0 0 ] [ 0 a million 0 ] [ 0 0 a million ] [ a million 0 0 ] [ 0 a million 0 ] [ 0 0 a million ] [ a million 0 0 ] [ 0 a million 0 ] and requiring that f to be a function skill that the intersection of C with any of the planes given via {x = consistent} {y = consistent} {z = consistent} has a single factor. this is, in some experience, the graphical representation of the scheme. f is now represented via the rotation matrix. We be conscious that we've constrained freedom in pick the curve C. A sufficiently accepted shape (that maximum unquestionably does not get well all a possibility purposes) is here: the matrix above generates a discrete subgroup G (with 3 aspects) of O(3), that fixes the subspace spanned via the vector (a million,a million,a million). to that end we are able to construct the orbifold O(3)/G, which we call the classic area. on the orbifold we are able to very almost freely prescribe a Curve: we merely might desire to make useful that the curve C defines a function. an undemanding thank you to make useful that C defines a function is to %. a representation of the orbifold (or a subset therefor) this is "good". working example, we are able to restrict to the slab { -infinity < x < 0, a (a,infty) g: (-infty,0) -> (0,a) Then the function f given via f (x) = h(x) if x < 0 g(h^{-a million}(x)) if x > a g^{-a million}(x) if 0

Answer 5

1) I'm not sure you wrote the first one correctly. I think it should be:
f(n+1) = [3*f(n) + 1 ] / 3
= f(n) + 1/3

So f(n+99) = f(n) + 99/3
f(100) = 35

2) f(x) = ax^2 + bx + c
We require that f(1) = 12 = a + b + c
Also f(x+1) = a*(x+1)^2 + b(x+1) + c
= ax^2 + bx + c + 2x + 8

We get:
ax^2 + (b+2a)*x + (a+b+c) = ax^2 + (b+2)*x + (c+8)

Equating coeffs:
a = 1, c+8 = 12, c = 4
b = 7

f(x) = x^2 + 7x + 4

3) The expression -ve when 1 of the two products is -ve and the other +ve.
Case 1:
|x| - 1 < 0 AND |x| + 2 > 0
which yields |x| < 1 or -1 < x < 1

Case 2:|x| - 1 > 0 AND |x| + 2 < 0
cannot occur

Ans. -1 < x < 1