All categories

3 years ago

Consider the exponential function f(x) = 2(3x) and its graph.

Mathematics

2 answers:

AveGali [126]3 years ago

6 0

Answer:

The initial value of the function is 2

The base of the function is 3

The function shows exponential growth

Step-by-step explanation:

f(x) = 2(3^x)

Exponential functions are those with the following equation:

y = a*b^x

where a ≠ 0, b > 0 and b ≠ 1 and x is a real number.

a is the y-intercept and b is the base.

The initial value of the function is the same as the y-intercept

If a is positive, the function growth. If a is negative, the function decay

geniusboy [140]3 years ago

6 0

Answer:

The initial value of the function is 2

The base of the function is 3

The function shows exponential growth

Step-by-step explanation:

You might be interested in

Find the measure of the central angle below: 45°

Semenov [28]

The central angle below is 1

4 0

2 years ago

What is m Help Please

olga_2 [115]

I don't know I forgot

3 0

2 years ago

Read 2 more answers

Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :