Question

Решите неравенство -16/(x^2-6x-7)<=0

Answer 1

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Answer 2

To solve the inequality $-\frac{16}{x^2-6x-7} \leq 0$, we need to find the values of $x$ that make the expression on the left-hand side non-positive (less than or equal to zero).

Step 1: Factorize the denominator The denominator $x^2-6x-7$ can be factored as $(x-7)(x+1)$.

Step 2: Identify critical points The critical points are the values of $x$ that make the expression undefined, which occur when the denominator is equal to zero. So, set $x^2-6x-7 = 0$ and solve for $x$: $(x-7)(x+1) = 0$

This gives us two critical points: $x = 7$ and $x = -1$.

Step 3: Create a sign chart Now, create a sign chart to determine the sign of the expression $-\frac{16}{x^2-6x-7}$ for different intervals of $x$:

$$\begin{array}{c|ccc|c} \textrm{Interval} & x<-1 & -1<x<7 & x>7 & \textrm{Critical Points} \\ \hline x+1 & - & + & + & x=-1 \\ x-7 & - & - & + & x=7 \\ -\frac{16}{x^2-6x-7} & ? & ? & ? & \\ \end{array}$$

Step 4: Test points in each interval Now, we need to pick test points in each interval and substitute them into the expression $-\frac{16}{x^2-6x-7}$ to determine the sign in each interval.

Let's test $x = -2$ (an arbitrary value less than -1): $-\frac{16}{(-2)^2-6(-2)-7} = -\frac{16}{4+12-7} = -\frac{16}{9} < 0$

Let's test $x = 0$ (an arbitrary value between -1 and 7): $-\frac{16}{0^2-6(0)-7} = -\frac{16}{-7} > 0$

Let's test $x = 8$ (an arbitrary value greater than 7): $-\frac{16}{8^2-6(8)-7} = -\frac{16}{64-48-7} = -\frac{16}{9} < 0$

Step 5: Determine the solution Now, we can fill in the sign chart:

$$\begin{array}{c|ccc|c} \textrm{Interval} & x<-1 & -1<x<7 & x>7 & \textrm{Critical Points} \\ \hline x+1 & - & + & + & x=-1 \\ x-7 & - & - & + & x=7 \\ -\frac{16}{x^2-6x-7} & + & - & + & \\ \end{array}$$

The inequality $-\frac{16}{x^2-6x-7} \leq 0$ is satisfied when $x \leq -1$ or $x \geq 7$. So, the solution to the inequality is the union of the intervals $(- \infty, -1]$ and $[7, \infty)$:

$-\infty < x \leq -1$ or $x \geq 7$.