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3 years ago

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If $1000 is invested at 8% interest, compounded annually, then after n years the investment is worthan = 1000(1.08)ndollars.(a)

Find the first five terms of the sequence {an}. (Round your answers to the nearest cent.)a1 = $a2 = $a3 = $a4 = $a5 = $(b) Is the sequence convergent or divergent?

1 answer:

Luba_88 [7]3 years ago

5 0

Answer:

Diverge

Step-by-step explanation:

(a)

1st year: $1000 * 1.08^1 = \$1080$

2nd year: $1000 * 1.08^2 = \$1166.4$

3rd year: $1000 * 1.08^3 = \$1259.71$

4th year: $1000 * 1.08^4 = \$1360.49$

5th year: $1000 * 1.08^5 = \$1469.33$

(b) The sequence is divergent, because if we take the derivative of the function with respect to n year:

$(1000*1.08^n)^{'} = 1000ln(n)*1.08^n$

This is a positive, meaning the slope of the function is positive. If we take the second derivative using product rule

$(1000ln(n)*1.08^n)^{'} = 1000\frac{ln(n)}{n}1.08^n$

This is also positive when n > 0. Therefore, the slope is positive and increasing. This means the sequence diverges.

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Answer the following question about the function whose derivative is given below

Komok [63]

Answer:

a) The critical points are $x = 3$ and $x = -6$ .

b) f is decreasing in the interval $(-\infty, -6)$

f is increasing in the intervals $(-6,3)$ and $(3,\infty)$ .

c) Local minima: $x = -6$

Local maxima: No local maxima

Step-by-step explanation:

(a) what are the critical points of f?

The critical points of f are those in which $f^{\prime}(x) = 0$ . So

$f^{\prime}(x) = 0$

$(x-3)^{2}(x+6) = 0$

So, the critical points are $x = 3$ and $x = -6$ .

(b) on what intervals is f increasing or decreasing? (if there is no interval put no interval)

For any interval, if $f^{\prime}$ is positive, f is increasing in the interval. If it is negative, f is decreasing in the interval.

Our critical points are $x = 3$ and $x = -6$ . So we have those following intervals:

$(-\infty, -6), (-6,3), (3, \infty)$

We select a point x in each interval, and calculate $f^{\prime}(x)$ .

So

-------------------------

$(-\infty, -6)$

$f^{\prime}(-7) = (-7-3)^{2}(-7+6) = (100)(-1) = -100$

f is decreasing in the interval $(-\infty, -6)$

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$(-6,3)$

$f^{\prime}(2) = (2-3)^{2}(2+6) = (1)(8) = 8$

f is increasing in the interval $(-6,3)$ .

------------------------------

$(3, \infty)$

$f^{\prime}(4) = (4-3)^{2}(4+6) = (1)(10) = 10$

f is increasing in the interval $(3,\infty)$ .

(c) At what points, if any, does f assume local maximum and minima values. ( if there is no local maxima put mo local maxima) if there is no local minima put no local minima

At a critical point x, if the function goes from decreasing to increasing, it is a local minima. And if the function goes from increasing to decreasing, it is a local maxima.

So, for each critical point is this problem:

At $x = -6$ , f goes from decreasing to increasing.

So $x = -6$ , f assume a local minima value

At $x = 3$ , f goes from increasing to increasing. So, there it is not a local maxima nor a local minima. So, there is no local maxima for this function.

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