Respuesta :
[tex]x \leq \dfrac{4x^2-13x}{x^2-9}[/tex]
Subtract both sides by [tex]\dfrac{4x^2-13x}{x^2-9}[/tex]
[tex]x- \dfrac{4x^2-13x}{x^2-9} \leq 0[/tex]
Simplify the left side into a single fraction. Note: [tex]x \rightarrow \dfrac{x^3-9x}{x^2-9} [/tex]
[tex]\dfrac{x^3-4x^2+4x}{x^2-9} \leq 0[/tex]
Factor the numerator and the denominator.
[tex]\dfrac{x(x-2)^2}{(x+3)(x-3)} \leq 0[/tex]
The equation on the left hand side has 2 zeroes and two places where it's undefined.
[tex]x=0, x=2[/tex]
These are the zeroes
[tex]x=3, x=-3[/tex]
These are where the function is undefined.
Using these values of x as "markers" or "checkpoints", we can create 5 possible solutions to this inequality. Using the equation in the left side of the inequality, we can test which possible solutions produce nonpositive (meaning it's negative or 0, but never positive) values.
Any intervals that produce nonpositve values for the left side of the inequality will be able to satisfy the inequality.
[tex]x\textless-3[/tex]
This produces nonpositive numbers. It's a solution. (Note: I didn't use the less than or equal sign because we don't want to have an undefined value included in the interval).
[tex]-3 \textless x \leq 0[/tex]
This produces nonnegative numbers. Not a solution.
(Note: I didn't use the less than or equal sign because we don't want to have an undefined value included in the interval).
[tex]0 \leq x \leq 2[/tex]
This produces nonpositive numbers. It's a solution.
[tex]2 \leq x \textless 3[/tex]
This gives more nonpositive numbers, it's a solution. Since both the previous and this intervals are solutions, they can be combined into [tex]0 \leq x \textless 3[/tex]
(Note: I didn't use the less than or equal sign because we don't want to have an undefined value included in the interval).
[tex]x \textgreater 3[/tex]
This produces positive values. It's not a solution.
(Note: I didn't use the greater than or equal sign because we don't want to have an undefined value included in the interval).
The solution is:
[tex]x \textless -3[/tex] or [tex]0 \leq x \textless 3[/tex]
Subtract both sides by [tex]\dfrac{4x^2-13x}{x^2-9}[/tex]
[tex]x- \dfrac{4x^2-13x}{x^2-9} \leq 0[/tex]
Simplify the left side into a single fraction. Note: [tex]x \rightarrow \dfrac{x^3-9x}{x^2-9} [/tex]
[tex]\dfrac{x^3-4x^2+4x}{x^2-9} \leq 0[/tex]
Factor the numerator and the denominator.
[tex]\dfrac{x(x-2)^2}{(x+3)(x-3)} \leq 0[/tex]
The equation on the left hand side has 2 zeroes and two places where it's undefined.
[tex]x=0, x=2[/tex]
These are the zeroes
[tex]x=3, x=-3[/tex]
These are where the function is undefined.
Using these values of x as "markers" or "checkpoints", we can create 5 possible solutions to this inequality. Using the equation in the left side of the inequality, we can test which possible solutions produce nonpositive (meaning it's negative or 0, but never positive) values.
Any intervals that produce nonpositve values for the left side of the inequality will be able to satisfy the inequality.
[tex]x\textless-3[/tex]
This produces nonpositive numbers. It's a solution. (Note: I didn't use the less than or equal sign because we don't want to have an undefined value included in the interval).
[tex]-3 \textless x \leq 0[/tex]
This produces nonnegative numbers. Not a solution.
(Note: I didn't use the less than or equal sign because we don't want to have an undefined value included in the interval).
[tex]0 \leq x \leq 2[/tex]
This produces nonpositive numbers. It's a solution.
[tex]2 \leq x \textless 3[/tex]
This gives more nonpositive numbers, it's a solution. Since both the previous and this intervals are solutions, they can be combined into [tex]0 \leq x \textless 3[/tex]
(Note: I didn't use the less than or equal sign because we don't want to have an undefined value included in the interval).
[tex]x \textgreater 3[/tex]
This produces positive values. It's not a solution.
(Note: I didn't use the greater than or equal sign because we don't want to have an undefined value included in the interval).
The solution is:
[tex]x \textless -3[/tex] or [tex]0 \leq x \textless 3[/tex]
