QuestionCompute the values of a periodic function $g(x)=\frac{x+1}{x^{2}-x+4}$ with period 7 on the interval $[-2,5)$ . Find all roots of $g(x)$ .

Studdy Solution

STEP 1

Assumptions
1. The function $g(x)$ is periodic with period 7.
2. The definition of $g(x)$ on the interval $[-2,5)$ is $g(x)=\frac{x+1}{x^{2}-x+4}$ .
3. The values of $g(x)$ outside the interval $[-2,5)$ can be found using the periodicity of the function.
4. To find the roots of $g(x)$ , we need to solve $g(x)=0$ within the interval $[-2,5)$ and then apply the periodicity.

STEP 2

To compute $g(5)$ , we need to find a value $x_0$ in the interval $[-2,5)$ such that $x_0$ is equivalent to 5 modulo 7 (the period of $g$ ).
$x_0 \equiv 5 \pmod{7}$

STEP 3

Since 5 is already in the interval $[-2,5)$ , we can directly substitute $x=5$ into the function $g(x)$ .
$g(5) = \frac{5+1}{5^{2}-5+4}$

STEP 4

Calculate $g(5)$ .
$g(5) = \frac{6}{25-5+4}$
$g(5) = \frac{6}{24}$
$g(5) = \frac{1}{4}$
(a) $g(5) = \frac{1}{4}$ .

STEP 5

To compute $g(42)$ , we need to find a value $x_1$ in the interval $[-2,5)$ such that $x_1$ is equivalent to 42 modulo 7.
$x_1 \equiv 42 \pmod{7}$

STEP 6

Since 42 is a multiple of 7, we can reduce it modulo 7 to get 0.
$x_1 \equiv 0 \pmod{7}$

STEP 7

Now we substitute $x=0$ into the function $g(x)$ because 0 is in the interval $[-2,5)$ .
$g(0) = \frac{0+1}{0^{2}-0+4}$

STEP 8

Calculate $g(0)$ .
$g(0) = \frac{1}{4}$
(b) $g(42) = g(0) = \frac{1}{4}$ .

STEP 9

To compute $g(-256)$ , we need to find a value $x_2$ in the interval $[-2,5)$ such that $x_2$ is equivalent to -256 modulo 7.
$x_2 \equiv -256 \pmod{7}$

STEP 10

Calculate $x_2$ by reducing -256 modulo 7.
$x_2 \equiv -256 \equiv 3 \pmod{7}$

STEP 11

Now we substitute $x=3$ into the function $g(x)$ because 3 is in the interval $[-2,5)$ .
$g(3) = \frac{3+1}{3^{2}-3+4}$

STEP 12

Calculate $g(3)$ .
$g(3) = \frac{4}{9-3+4}$
$g(3) = \frac{4}{10}$
$g(3) = \frac{2}{5}$
(c) $g(-256) = g(3) = \frac{2}{5}$ .

STEP 13

To find all roots of $g$ , we need to solve the equation $g(x)=0$ within the interval $[-2,5)$ .
$g(x) = \frac{x+1}{x^{2}-x+4} = 0$

STEP 14

For the fraction to be zero, the numerator must be zero (as the denominator cannot be zero because it is a quadratic with no real roots).
$x+1 = 0$

STEP 15

Solve for $x$ .
$x = -1$

STEP 16

Check if $x=-1$ is in the interval $[-2,5)$ .
Since $-1$ is in the interval, it is a root of $g(x)$ .

STEP 17

Because $g(x)$ is periodic with period 7, all other roots can be found by adding multiples of 7 to $-1$ .
$x = -1 + 7k, \text{ where } k \text{ is an integer}$
(d) The roots of $g$ are $x = -1 + 7k$ for all integers $k$ .

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