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2 years ago

14

Find f(x) and g(x) so that the function can be described as y = f(g(x)). y = nine divided by square root of quantity five x plus

five.

2 answers:

katrin [286]2 years ago

8 0

$y=\frac{9}{\sqrt{5x+5}}$
hmm, seems that we can say that the square root thing is the g(x)

I would say
$f(x)=\frac{9}{x}$ and
$g(x)=\sqrt{5x+5}$

we could also have
$f(x)=\frac{9}{\sqrt{x}}$ and
$g(x)=5x+5$

or
$f(x)=\frac{9}{\sqrt{x+5}}$ and
$g(x)=5x$

there are many posibilities

Thepotemich [5.8K]2 years ago

8 0

Answer:

f(x)= $\frac{9}{x}$

g(x)= $\sqrt{5x+5}$

Step-by-step explanation:

y= $\frac{9}{\sqrt{5x+5} }$

We have to find f(x) and g(x). There are many possible answers, but we are going to suppose that f(x)= $\frac{9}{x}$

to obtain y(x) we should have $\sqrt{5x+5}$ in the denominator. We can obtain that by saying that g(x)= $\sqrt{5x+5}$ so f(g(x))= $\frac{9}{sqrt{5x+5}}$

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Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

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(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

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$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
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So we expect a general solution of the form

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With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

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$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
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