La función lineal f(x) que verifica f(−2) = 6 y f(1) = 3 es f(x) =

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Here are the solutions to the math problems:

1. The linear function $f(x)$ is of the form $f(x) = mx+b$ . We are given $f(-2)=6$ and $f(1)=3$ .

Use the given points to form a system of equations.
For $f(-2)=6$ :
$m(-2) + b = 6 \implies -2m + b = 6 \quad (1)$
For $f(1)=3$ :
$m(1) + b = 3 \implies m + b = 3 \quad (2)$
Solve the system of equations.
Subtract equation (2) from equation (1):
$(-2m + b) - (m + b) = 6 - 3$ $-3m = 3$ $m = -1$
Substitute $m=-1$ into equation (2) to find $b$ .
$-1 + b = 3$ $b = 4$
Write the function $f(x)$ .
$f(x) = -x + 4$
The correct option is $\boxed{-x+4}$ .

2. The vertex of the graph of $f(x) = 4x(x-2)$ is required.
Expand the function to the standard quadratic form $ax^2+bx+c$ .
$f(x) = 4x^2 - 8x$
Here, $a=4$ , $b=-8$ , $c=0$ .
Find the x-coordinate of the vertex using the formula $x_v = -\frac{b}{2a}$ .
$x_v = -\frac{-8}{2(4)} = \frac{8}{8} = 1$
Find the y-coordinate of the vertex by substituting $x_v=1$ into the function.
$y_v = f(1) = 4(1)(1-2) = 4(-1) = -4$
The vertex is $(1, -4)$ .
The correct option is $\boxed{(1,-4)}$ .

3. The set $A = \{x \in \mathbb{R} / \frac{x+3}{x} > 0\}$ requires the fraction to be positive. This happens when both numerator and denominator have the same sign.
Find the critical points where the numerator or denominator is zero.
$x+3 = 0 \implies x = -3$
$x = 0$
Create a sign chart using the critical points $-3$ and 0.
- For $x < -3$ (e.g., $x=-4$ ): $\frac{-4+3}{-4} = \frac{-1}{-4} = \frac{1}{4} > 0$ . So, $(-\infty, -3)$ is part of the solution.
- For $-3 < x < 0$ (e.g., $x=-1$ ): $\frac{-1+3}{-1} = \frac{2}{-1} = -2 < 0$ .
- For $x > 0$ (e.g., $x=1$ ): $\frac{1+3}{1} = \frac{4}{1} = 4 > 0$ . So, $(0, +\infty)$ is part of the solution.
Combine the intervals where the expression is positive.
The solution set is $(-\infty, -3) \cup (0, +\infty)$ .
The correct option is $\boxed{(-\infty;-3) \cup (0;+\infty)}$ .

4. The domain of $f(x) = \ln(4x-8)$ is found by ensuring the argument of the natural logarithm is strictly positive.
Set the argument of the logarithm greater than zero.
$4x - 8 > 0$
Solve for $x$ .
$4x > 8$ $x > \frac{8}{4}$ $x > 2$
The domain is $(2, +\infty)$ .
The correct option is $\boxed{(2;+\infty)}$ .

La función lineal f(x) que verifica f(-2) = 6 y f(1) = 3 es f(x) =

Step-by-step solution

Use the given points to form a system of equations.

Solve the system of equations.

Substitute $m=-1$ into equation (2) to find $b$ .

Write the function $f(x)$ .

Expand the function to the standard quadratic form $ax^2+bx+c$ .

Find the x-coordinate of the vertex using the formula $x_v = -\frac{b}{2a}$ .

Find the y-coordinate of the vertex by substituting $x_v=1$ into the function.

Find the critical points where the numerator or denominator is zero.

Create a sign chart using the critical points $-3$ and 0.

Combine the intervals where the expression is positive.

Set the argument of the logarithm greater than zero.

Solve for $x$ .

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More Mathematics Questions

La función lineal f(x) que verifica f(-2) = 6 y f(1) = 3 es f(x) =

Step-by-step solution

Use the given points to form a system of equations.

Solve the system of equations.

Substitute m=−1m=-1m=−1 into equation (2) to find bbb.

Write the function f(x)f(x)f(x).

Expand the function to the standard quadratic form ax2+bx+cax^2+bx+cax2+bx+c.

Find the x-coordinate of the vertex using the formula xv=−b2ax_v = -\frac{b}{2a}xv​=−2ab​.

Find the y-coordinate of the vertex by substituting xv=1x_v=1xv​=1 into the function.