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Veseljchak [2.6K]

3 years ago

8

Write an equation for the graph below, which represents an exponential function f with base 2 or 3, translated and/or reflect

ed.

1 answer:

evablogger [386]3 years ago

6 0

Answer:

$f(x) = 2*2^x + 3$

Step-by-step explanation:

The model for an exponencial function is:

$f(x) = a*b^x + c$

'c' is the value of the asymptote, so we have c = 3.

To find the value of 'a', let's use the point (0,5):

$5 = a*b^0 + 3$

$5 = a*1 + 3$

$a = 2$

Now, to find the value of 'b', let's use the point (1, 7):

$7 = 2 * b^1 + 3$

$7 = 2b + 3$

$2b = 4$

$b = 2$

So our function is:

$f(x) = 2*2^x + 3$

Let's check if the point (-1, 4) is inside our function:

$4 = 2 * 2^{-1} + 3$

$4 = 1 + 3$

$4 = 4\ (correct)$

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AfilCa [17]

<h3>4 Answers: A, B, C, D</h3>

=======================================================

Explanation:

f(x) is continuous when x >= 1. The only discontinuity for f(x) is when x = 0, but 0 is not part of this interval.

f(x) is positive for any valid x value in the domain since x^6 is always positive. In general, x^n is positive for all x when n is any even number.

f(x) is decreasing. You can see this through a table of values or through a graph. For anything in the form 1/(x^k), it will be a decreasing function because x^k gets larger, so 1/(x^k) gets smaller, when x goes to infinity.

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

The conditions to use the integral test have been met. So we have to see if $\displaystyle \int_1^{\infty}f(x)dx$ converges or not.

Let's integrate and find out

$\displaystyle \int \frac{1}{x^6} dx = \int x^{-6} dx\\\\\\ \displaystyle \int \frac{1}{x^6} dx = \frac{1}{1+(-6)}x^{-6+1}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = \frac{1}{-5}x^{-5}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = -\frac{1}{5}*\frac{1}{x^5}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = -\frac{1}{5x^5}+C\\\\$

So we have

$\displaystyle g(x) = \int f(x) dx\\\\\\\displaystyle g(x) = \int \frac{1}{x^6} dx\\\\\\\displaystyle g(x) = -\frac{1}{5x^5}+C\\\\\\$

Meaning that,

$\displaystyle \int_{a}^{b} f(x) dx = g(b)-g(a)\\\\\\\displaystyle \int_{a}^{b} \frac{1}{x^6} dx = \left(-\frac{1}{5b^5}+C\right)-\left(-\frac{1}{5a^5}+C\right)\\\\\\\displaystyle \int_{a}^{b} \frac{1}{x^6} dx = -\frac{1}{5b^5}+\frac{1}{5a^5}\\\\\\$

If we plug in a = 1 and apply the limit as b approaches positive infinity, then the expression $-\frac{1}{5b^5}+\frac{1}{5a^5}$ will turn into $\frac{1}{5}$

Therefore,

$\displaystyle \int_{1}^{\infty} \frac{1}{x^6} dx = \frac{1}{5}\\\\\\$

Because this integral converges, this means the series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^6}$ also converges as well by the integral test.

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3 years ago